Download pdf - Commutative Algebra - TU Kaiserslauterngathmann/class/... · Commutative algebra is the study of commutative rings. In this class we will assume the basics of ring theory that you

Commutative AlgebraAndreas Gathmann

Class Notes TU Kaiserslautern 2013/14

Contents

0. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 31. Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 92. Prime and Maximal Ideals. . . . . . . . . . . . . . . . . . . . . 183. Modules . . . . . . . . . . . . . . . . . . . . . . . . . . 274. Exact Sequences . . . . . . . . . . . . . . . . . . . . . . . . 365. Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . 436. Localization . . . . . . . . . . . . . . . . . . . . . . . . . 527. Chain Conditions. . . . . . . . . . . . . . . . . . . . . . . . 628. Prime Factorization and Primary Decompositions . . . . . . . . . . . . . 709. Integral Ring Extensions . . . . . . . . . . . . . . . . . . . . . 80

10. Noether Normalization and Hilbert’s Nullstellensatz . . . . . . . . . . . . 9111. Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 9612. Valuation Rings . . . . . . . . . . . . . . . . . . . . . . . . 10913. Dedekind Domains . . . . . . . . . . . . . . . . . . . . . . . 117

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

0. Introduction 3

0. Introduction

Commutative algebra is the study of commutative rings. In this class we will assume the basicsof ring theory that you already know from earlier courses (e. g. ideals, quotient rings, the homo-morphism theorem, and unique prime factorization in principal ideal domains such as the integersor polynomial rings in one variable over a field), and move on to more advanced topics, some ofwhich will be sketched in Remark 0.14 below. For references to earlier results I will usually use myGerman notes for the “Algebraic Structures” and occasionally the “Foundations of Mathematics”and “Introduction to Algebra” classes [G1, G2, G3], but if you prefer English references you willcertainly have no problems to find them in almost any textbook on abstract algebra.

You will probably wonder why the single algebraic structure of commutative rings deserves a fullone-semester course for its study. The main motivation for this is its many applications in bothalgebraic geometry and (algebraic) number theory. Especially the connection between commutativealgebra and algebraic geometry is very deep — in fact, to a certain extent one can say that these twofields of mathematics are essentially the same thing, just expressed in different languages. Althoughsome algebraic constructions and results in this class may seem a bit abstract, most of them have aneasy (and sometimes surprising) translation in terms of geometry, and knowing about this often helpsto understand and remember what is going on. For example, we will see that the Chinese RemainderTheorem that you already know [G1, Proposition 11.22] (and that we will extend to more generalrings than the integers in Proposition 1.14) can be translated into the seemingly obvious geometricstatement that “giving a function on a disconnected space is the same as giving a function on eachof its connected components” (see Example 1.15 (b)).

However, as this is not a geometry class, we will often only sketch the correspondence between al-gebra and geometry, and we will never actually use algebraic geometry to prove anything. Althoughour “Commutative Algebra” and “Algebraic Geometry” classes are deeply linked, they are deliber-ately designed so that none of them needs the other as a prerequisite. But I will always try to giveyou enough examples and background to understand the geometric meaning of what we do, in caseyou have not attended the “Algebraic Geometry” class yet.

So let us explain in this introductory chapter how algebra enters the field of geometry. For thiswe have to introduce the main objects of study in algebraic geometry: solution sets of polynomialequations over some field, the so-called varieties.

Convention 0.1 (Rings and fields). In our whole course, a ring R is always meant to be a commu-tative ring with 1 [G1, Definition 7.1]. We do not require that this multiplicative unit 1 is distinctfrom the additive neutral element 0, but if 1 = 0 then R must be the zero ring [G1, Lemma 7.5 (c)].Subrings must have the same unit as the ambient ring, and ring homomorphisms are always requiredto map 1 to 1. Of course, a ring R 6= {0} is a field if and only if every non-zero element has amultiplicative inverse.

Definition 0.2 (Polynomial rings). Let R be a ring, and let n ∈ N>0. A polynomial over R in nvariables is a formal expression of the form

f = ∑i1,...,in∈N

ai1,...,in xi11 · · · · · x

inn ,

with coefficients ai1,...,in ∈ R and formal variables x = (x1, . . . ,xn), such that only finitely many of thecoefficients are non-zero (see [G1, Chapter 9] how this concept of “formal variables” can be definedin a mathematically rigorous way).

Polynomials can be added and multiplied in the obvious way, and form a ring with these operations.We call it the polynomial ring over R in n variables and denote it by R[x1, . . . ,xn].

4 Andreas Gathmann

Definition 0.3 (Varieties). Let K be a field, and let n ∈ N.

(a) We callAn

K := {(c1, . . . ,cn) : ci ∈ K for i = 1, . . . ,n}the affine n-space over K. If the field K is clear from the context, we will write An

K also asAn.

Note that AnK is just Kn as a set. It is customary to use two different notations here since

Kn is also a K-vector space and a ring. We will usually use the notation AnK if we want to

ignore these additional structures: for example, addition and scalar multiplication are definedon Kn, but not on An

K . The affine space AnK will be the ambient space for our zero loci of

polynomials below.

(b) For a polynomial f ∈ K[x1, . . . ,xn] as above and a point c = (c1, . . . ,cn) ∈ AnK we define the

value of f at c to be

f (c) = ∑i1,...,in∈N

ai1,...,in ci11 · · · · · c

inn ∈ K.

If there is no risk of confusion we will sometimes denote a point in AnK by the same letter

x as we used for the formal variables, writing f ∈ K[x1, . . . ,xn] for the polynomial and f (x)for its value at a point x ∈ An

K .

(c) Let S⊂ K[x1, . . . ,xn] be a set of polynomials. Then

V (S) := {x ∈ AnK : f (x) = 0 for all f ∈ S} ⊂ An

K

is called the zero locus of S. Subsets of AnK of this form are called (affine) varieties. If

S = ( f1, . . . , fk) is a finite set, we will write V (S) =V ({ f1, . . . , fk}) also as V ( f1, . . . , fk).

Example 0.4. Varieties, say over the field R of real numbers, can have many different “shapes”.The following picture shows a few examples in A2

R and A3R.

(a) V (x21 + x2

2−1)⊂ A2 (b) V (x22− x3

1)⊂ A2 (c) V (x31− x1)⊂ A2

(d) V (x61 + x6

2 + x63−1)⊂ A3 (e) V (x1x3,x2x3)⊂ A3 (f) V (x2

2 + x33− x4

3− x21x2

3)⊂ A3

Of course, the empty set /0 and all of An are also varieties in An, since /0 =V (1) and An =V (0).

It is the goal of algebraic geometry to find out the geometric properties of varieties by lookingat the corresponding polynomials from an algebraic point of view (as opposed to an analytical ornumerical approach). However, it turns out that it is not a very good idea to just look at the definingpolynomials given initially — simply because they are not unique. For example, the variety (a)above was given as the zero locus of the polynomial x2

1 + x22−1, but it is equally well the zero locus

of (x21 + x2

2− 1)2, or of the two polynomials (x1− 1)(x21 + x2

2− 1) and x2 (x21 + x2

2− 1). In order to

0. Introduction 5

remove this ambiguity, it is therefore useful to consider all polynomials vanishing on X at once. Letus introduce this concept now.

Construction 0.5 (Rings and ideals associated to varieties). For a variety X ⊂ AnK (and in fact also

for any subset X of AnK) we consider the set

I(X) := { f ∈ K[x1, . . . ,xn] : f (x) = 0 for all x ∈ X}of all polynomials vanishing on X . Note that this is an ideal of K[x1, . . . ,xn] (which we write asI(X)EK[x1, . . . ,xn]): it is clear that 0 ∈ I(X), and if two polynomials f and g vanish on X , then sodo f +g and f ·h for any polynomial h. We call I(X) the ideal of X .

With this ideal we can construct the quotient ring

A(X) := K[x1, . . . ,xn]/I(X)

in which we identify two polynomials f ,g ∈ K[x1, . . . ,xn] if and only if f −g is the zero function onX , i. e. if f and g have the same value at every point x∈ X . So one may think of an element f ∈ A(X)as being the same as a function

X → K, x 7→ f (x)that can be given by a polynomial. We therefore call A(X) the ring of polynomial functions orcoordinate ring of X . Often we will simply say that A(X) is the ring of functions on X sincefunctions in algebra are always given by polynomials. Moreover, the class of a polynomial f ∈K[x1, . . . ,xn] in such a ring will usually also be written as f ∈ A(X), dropping the explicit notationfor equivalence classes if it is clear from the context that we are talking about elements in the quotientring.

Remark 0.6 (Polynomials and polynomial functions). You probably know that over some fieldsthere is a subtle difference between polynomials and polynomial functions: e. g. over the field K =Z2 the polynomial f = x2 + x ∈ K[x] is certainly non-zero, but it defines the zero function on A1

K[G1, Remark 9.16 (b)]. In our current notation this means that the ideal I(A1

K) of functions vanishingat every point of A1

K is non-trivial, in fact that I(A1K) = (x2 + x), and that consequently the ring

A(A1K) = K[x]/(x2 + x) of polynomial functions on A1

K is not the same as the polynomial ring K[x].

In this class we will skip over this problem entirely, since our main geometric intuition comes fromthe fields of real or complex numbers where there is no difference between polynomials and polyno-mial functions. We will therefore usually assume silently that there is no polynomial f ∈K[x1, . . . ,xn]vanishing on all of An

K , i. e. that I(AnK) = (0) and thus A(An

K) = K[x1, . . . ,xn].

Example 0.7 (Ideal of a point). Let a = (a1, . . . ,an) ∈AnK be a point. We claim that its ideal I(a) :=

I({a})EK[x1, . . . ,xn] isI(a) = (x1−a1, . . . ,xn−an).

In fact, this is easy to see:

“⊂” If f ∈ I(a) then f (a) = 0. This means that replacing each xi by ai in f gives zero, i. e. that fis zero modulo (x1−a1, . . . ,xn−an). Hence f ∈ (x1−a1, . . . ,xn−an).

“⊃” If f ∈ (x1−a1, . . . ,xn−an) then f = ∑ni=1(xi−ai) fi for some f1, . . . , fn ∈ K[x1, . . . ,xn], and

so certainly f (a) = 0, i. e. f ∈ I(a).

Construction 0.8 (Subvarieties). The ideals of varieties defined in Construction 0.5 all lie in thepolynomial ring K[x1, . . . ,xn]. In order to get a geometric interpretation of ideals in more generalrings it is useful to consider a relative situation: let X ⊂ An

K be a fixed variety. Then for any subsetS⊂ A(X) of polynomial functions on X we can consider its zero locus

VX (S) = {x ∈ X : f (x) = 0 for all f ∈ S} ⊂ X

just as in Definition 0.3 (c), and for any subset Y ⊂ X as in Construction 0.5 the ideal

IX (Y ) = { f ∈ A(X) : f (x) = 0 for all x ∈ Y} EA(X)

of all functions on X that vanish on Y . It is clear that the sets of the form VX (S) are exactly thevarieties in An

K contained in X , the so-called subvarieties of X .

6 Andreas Gathmann

If there is no risk of confusion we will simply write V (S) and I(Y ) again instead of VX (S) and IX (Y ).So in this notation we have now assigned to every variety X a ring A(X) of polynomial functionson X , and to every subvariety Y ⊂ X an ideal I(Y )EA(X) of the functions that vanish on Y . Thisassignment of an ideal to a subvariety has some nice features:

Lemma 0.9. Let X be a variety with coordinate ring A(X). Moreover, let Y and Y ′ be subsets of X,and let S and S′ be subsets of A(X).

(a) If Y ⊂ Y ′ then I(Y ′)⊂ I(Y ) in A(X); if S⊂ S′ then V (S′)⊂V (S) in X.

(b) Y ⊂V (I(Y )) and S⊂ I(V (S)).

(c) If Y is a subvariety of X then Y =V (I(Y )).

(d) If Y is a subvariety of X then A(X)/I(Y )∼= A(Y ).

Proof.

(a) Assume that Y ⊂ Y ′. If f ∈ I(Y ′) then f vanishes on Y ′, hence also on Y , which means thatf ∈ I(Y ). The second statement follows in a similar way.

(b) Let x ∈ Y . Then f (x) = 0 for every f ∈ I(Y ) by definition of I(Y ). But this implies thatx ∈V (I(Y )). Again, the second statement follows analogously.

(c) By (b) it suffices to prove “⊃”. As Y is a subvariety of X we can write Y = V (S) for someS⊂ A(X). Then S⊂ I(V (S)) by (b), and thus V (S)⊃V (I(V (S))) by (a). Replacing V (S) byY now gives the required inclusion.

(d) The ring homomorphism A(X)→A(Y ) that restricts a polynomial function on X to a functionY is surjective and has kernel I(Y ) by definition. So the result follows from the homomor-phism theorem [G1, Proposition 8.12]. �

Remark 0.10 (Reconstruction of geometry from algebra). Let Y be a subvariety of X . Then Lemma0.9 (c) says that I(Y ) determines Y uniquely. Similarly, knowing the rings A(X) and A(Y ), togetherwith the ring homomorphism A(X)→ A(Y ) that describes the restriction of functions on X to func-tions on Y , is enough to recover I(Y ) as the kernel of this map, and thus Y as a subvariety of X bythe above. In other words, we do not lose any information if we pass from geometry to algebra anddescribe varieties and their subvarieties by their coordinate rings and ideals.

This map A(X)→ A(Y ) corresponding to the restriction of functions to a subvariety is already a firstspecial case of a ring homomorphism associated to a “morphism of varieties”. Let us now introducethis notion.

Construction 0.11 (Morphisms of varieties). Let X ⊂ AnK and Y ⊂ Am

K be two varieties over thesame ground field. Then a morphism from X to Y is just a set-theoretic map f : X → Y that canbe given by polynomials, i. e. such that there are polynomials f1, . . . , fm ∈ K[x1, . . . ,xn] with f (x) =( f1(x), . . . , fm(x)) ∈ Y for all x ∈ X . To such a morphism we can assign a ring homomorphism

ϕ : A(Y )→ A(X), g 7→ g◦ f = g( f1, . . . , fm)

given by composing a polynomial function on Y with f to obtain a polynomial function on X . Notethat this ring homomorphism ϕ . . .

(a) reverses the roles of source and target compared to the original map f : X → Y ; and

(b) is enough to recover f , since fi = ϕ(yi) ∈ A(X) if y1, . . . ,ym denote the coordinates of AmK .

Example 0.12. Let X = A1R (with coordinate x) and Y = A2

R (with coordinates y1 and y2), so thatA(X) = R[x] and A(Y ) = R[y1,y2] by Remark 0.6. Consider the morphism of varieties

f : X → Y, x 7→ (y1,y2) := (x,x2)

0. Introduction 7

whose image is obviously the standard parabola Z =V (y2−y21) shown in

the picture on the right. Then the associated ring homomorphism A(Y ) =R[y1,y2]→ R[x] = A(X) of Construction 0.11 is given by composing apolynomial function in y1 and y2 with f , i. e. by plugging in x and x2 fory1 and y2, respectively:

R[y1,y2]→ R[x], g 7→ g(x,x2).

Note that with the images of g = y1 and g = y2 under this homomorphismwe just recover the polynomials x and x2 defining the map f .

If we had considered f as a morphism from X to Z (i. e. restricted thetarget space to the actual image of f ) we would have obtained A(Z) =K[y1,y2]/(y2− y2

1) and thus the ring homomorphism

f

Z

y2

y1

x

R[y1,y2]/(y2− y21)→ R[x], g 7→ g(x,x2)

instead (which is obviously well-defined).

Remark 0.13 (Correspondence between geometry and algebra). Summarizing what we have seenso far, we get the following first version of a dictionary between geometry and algebra:

GEOMETRY −→ ALGEBRAvariety X ring A(X) of (polynomial) functions on Xsubvariety Y of X ideal I(Y )EA(X) of functions on X vanishing on Ymorphism f : X → Y of varieties ring homomorphism A(Y )→ A(X), g 7→ g◦ f

Moreover, passing from ideals to subvarieties reverses inclusions as in Lemma 0.9 (a), and we haveA(X)/I(Y ) ∼= A(Y ) for any subvariety Y of X by Lemma 0.9 (d) (with the isomorphism given byrestricting functions from X to Y ).

We have also seen already that this assignment of algebraic to geometric objects is injective in thesense of Remark 0.10 and Construction 0.11 (b). However, not all rings, ideals, and ring homo-morphisms arise from this correspondence with geometry, as we will see in Remark 1.10, Example1.25 (b), and Remark 1.31. So although the geometric picture is very useful to visualize algebraicstatements, it can usually not be used to actually prove them in the case of general rings.

Remark 0.14 (Outline of this class). In order to get an idea of the sort of problems considered incommutative algebra, let us quickly list some of the main topics that we will discuss in this class.

• Modules. From linear algebra you know that one of the most important structures relatedto a field K is that of a vector space over K. If we write down the same axioms as for avector space but relax the condition on K to allow an arbitrary ring, we obtain the algebraicstructure of a module, which is equally important in commutative algebra as that of a vectorspace in linear algebra. We will study this in Chapter 3.

• Localization. If we have a ring R that is not a field, an important construction discussed inChapter 6 is to make more elements invertible by allowing “fractions” — in the same wayas one can construct the rational numbers Q from the integers Z. Geometrically, we will seethat this process corresponds to studying a variety locally around a point, which is why it iscalled “localization”.

• Decomposition into primes. In a principal ideal domain R like the integers or a polynomialring in one variable over a field, an important algebraic tool is the unique prime factorizationof elements of R [G1, Proposition 11.9]. We will extend this concept in Chapter 8 to moregeneral rings, and also to a “decomposition of ideals into primes”. In terms of geometry,this corresponds to a decomposition of a variety into pieces that cannot be subdivided anyfurther — e. g. writing the variety in Example 0.4 (e) as a union of a line and a plane.

8 Andreas Gathmann

• Dimension. Looking at Example 0.4 again it seems obvious that we should be able to assigna dimension to each variety X . We will do this by assigning a dimension to each commutativering so that the dimension of the coordinate ring A(X) can be interpreted as the geometricdimension of X (see Chapter 11). With this definition of dimension we can then proveits expected properties, e. g. that cutting down a variety by n more equations reduces itsdimension by at most n (Remark 11.18).

• Orders of vanishing. For a polynomial f ∈ K[x] in one variable you all know what it meansthat it has a zero of a certain order at a point. If we now have a different variety, say stilllocally diffeomorphic to a line such as e. g. the circle X =V (x2

1 + x22−1)⊂ A2

R in Example0.4 (a), it seems geometrically reasonable that we should still be able to define such vanishingorders of functions on X at a given point. This is in fact possible, but algebraically morecomplicated — we will do this in Chapter 12 and study the consequences in Chapter 13.

But before we can discuss these main topics of the class we have to start now by developing moretools to work with ideals than what you know from earlier classes.

Exercise 0.15. Show that the following subsets X of AnK are not varieties over K:

(a) X = Z⊂ A1R;

(b) X = A1R\{0} ⊂ A1

R;

(c) X = {(x1,x2) ∈ A2R : x2 = sin(x1)} ⊂ A2

R;

(d) X = {x ∈ A1C : |x|= 1} ⊂ A1

C;

(e) X = f (Y )⊂ AnR for an arbitrary variety Y and a morphism of varieties f : Y → X over R.

Exercise 0.16 (Degree of polynomials). Let R be a ring. Recall that an element a ∈ R is called azero-divisor if there exists an element b 6= 0 with ab = 0 [G1, Definition 7.6 (c)], and that R is calledan (integral) domain if no non-zero element is a zero-divisor, i. e. if ab = 0 for a,b∈ R implies a = 0or b = 0 [G1, Definition 7.6 (d)].

We define the degree of a non-zero polynomial f = ∑i1,...,in ai1,...,in xi11 · · · · · xin

n ∈ R[x1, . . . ,xn] to be

deg f := max{i1 + · · ·+ in : ai1,...,in 6= 0}.Moreover, the degree of the zero polynomial is formally set to −∞. Show that:

(a) deg( f ·g)≤ deg f +degg for all f ,g ∈ R[x1, . . . ,xn].

(b) Equality holds in (a) for all polynomials f and g if and only if R is an integral domain.

1. Ideals 9

1. Ideals

From the “Algebraic Structures” class you already know the basic constructions and properties con-cerning ideals and their quotient rings [G1, Chapter 8]. For our purposes however we have to studyideals in much more detail — so this will be our goal for this and the next chapter. Let us start withsome general constructions to obtain new ideals from old ones. The ideal generated by a subset Mof a ring [G1, Definition 8.5] will be written as (M).

Construction 1.1 (Operations on ideals). Let I and J be ideals in a ring R.

(a) The sum of the two given ideals is defined as usual by

I + J := {a+b : a ∈ I and b ∈ J}.It is easy to check that this is an ideal — in fact, it is just the ideal generated by I∪ J.

(b) It is also obvious that the intersection I∩ J is again an ideal of R.

(c) We define the product of I and J as the ideal generated by all products of elements of I andJ, i. e.

I · J := ({ab : a ∈ I and b ∈ J}).Note that just the set of products of elements of I and J would in general not be an ideal: ifwe take R = R[x,y] and I = J = (x,y), then obviously x2 and y2 are products of an elementof I with an element of J, but their sum x2 + y2 is not.

(d) The quotient of I by J is defined to be

I :J := {a ∈ R : aJ ⊂ I}.Again, it is easy to see that this is an ideal.

(e) We call √I := {a ∈ R : an ∈ I for some n ∈ N}

the radical of I. Let us check that this an ideal of R:

• We have 0 ∈√

I, since 0 ∈ I.

• If a,b ∈√

I, i. e. an ∈ I and bm ∈ I for some n,m ∈ N, then

(a+b)n+m =n+m

∑k=0

(n+m

k

)ak bn+m−k

is again an element of I, since in each summand we must have that the power of a is atleast n (in which case ak ∈ I) or the power of b is at least m (in which case bn+m−k ∈ I).Hence a+b ∈

√I.

• If r ∈ R and a ∈√

I, i. e. an ∈ I for some n ∈ N, then (ra)n = rn an ∈ I, and hencera ∈

√I.

Note that we certainly have√

I ⊃ I. We call I a radical ideal if√

I = I, i. e. if for all a ∈ Rand n ∈ N with an ∈ I it follows that a ∈ I. This is a natural definition since the radical

√I

of an arbitrary ideal I is in fact a radical ideal in this sense: if an ∈√

I for some n, so anm ∈ Ifor some m, then this obviously implies a ∈

√I. 01

Whether an ideal I is radical can also easily be seen from its quotient ring R/I as follows.

Definition 1.2 (Nilradical, nilpotent elements, and reduced rings). Let R be a ring. The ideal√(0) = {a ∈ R : an = 0 for some n ∈ N}

is called the nilradical of R; its elements are called nilpotent. If R has no nilpotent elements except0, i. e. if the zero ideal is radical, then R is called reduced.

10 Andreas Gathmann

Lemma 1.3. An ideal IER is radical if and only if R/I is reduced.

Proof. By Construction 1.1 (e), the ideal I is radical if and only if for all a ∈ R and n ∈N with an ∈ Iit follows that a ∈ I. Passing to the quotient ring R/I, this is obviously equivalent to saying thata n = 0 implies a = 0, i. e. that R/I has no nilpotent elements except 0. �

Example 1.4 (Operations on ideals in principal ideal domains). Recall that a principal ideal do-main (or short: PID ) is an integral domain in which every ideal is principal, i. e. can be generatedby one element [G1, Definition 10.11]. The most prominent examples of such rings are probablyEuclidean domains, i. e. integral domains admitting a division with remainder [G1, Definition 10.17and Proposition 10.21], such as Z or K[x] for a field K [G1, Example 10.18 and Proposition 10.19].

We know that any principal ideal domain R admits a unique prime factorization of its elements [G1,Proposition 11.9] — a concept that we will discuss in more detail in Chapter 8. As a consequence,all operations of Construction 1.1 can then be computed easily: if I and J are not the zero ideal wecan write I = (a) and J = (b) for a= pa1

1 · · · · · pann and b= pb1

1 · · · · · pbnn with distinct prime elements

p1, . . . , pn and a1, . . . ,an,b1, . . . ,bn ∈ N. Then we obtain:

(a) I+J = (pc11 · · · · · pcn

n ) with ci =min(ai,bi) for i= 1, . . . ,n: another (principal) ideal containsI (resp. J) if and only if it is of the form (pc1

1 · · · · · pcnn ) with ci ≤ ai (resp. ci ≤ bi) for all i,

so the smallest ideal I + J containing I and J is obtained for ci = min(ai,bi);

(b) I∩ J = (pc11 · · · · · pcn

n ) with ci = max(ai,bi);

(c) I · J = (ab) = (pc11 · · · · · pcn

n ) with ci = ai +bi;

(d) I :J = (pc11 · · · · · pcn

n ) with ci = max(ai−bi,0);

(e)√

I = (pc11 · · · · · pcn

n ) with ci = min(ai,1).

In particular, we have I + J = (1) = R if and only if a and b have no common prime factor, i. e. if aand b are coprime. We use this observation to define the notion of coprime ideals in general rings:

Definition 1.5 (Coprime ideals). Two ideals I and J in a ring R are called coprime if I + J = R.

Example 1.6 (Operations on ideals in polynomial rings with SINGULAR). In more general rings,the explicit computation of the operations of Construction 1.1 is quite complicated and requiresadvanced algorithmic methods that you can learn about in the “Computer Algebra” class. We willnot need this here, but if you want to compute some examples in polynomial rings you can usee. g. the computer algebra system SINGULAR [S]. For example, for the ideals I = (x2y,xy3) andJ = (x+ y) in Q[x,y] the following SINGULAR code computes that I : J = (x2y,xy2) and

√I · J =√

I∩ J = (x2y+ xy2), and checks that y3 ∈ I + J:

> LIB "primdec.lib"; // library needed for the radical> ring R=0,(x,y),dp; // set up polynomial ring Q[x,y]> ideal I=x2y,xy3; // means I=(x^2*y,x*y^3)> ideal J=x+y;> quotient(I,J); // compute (generators of) I:J_[1]=xy2_[2]=x2y> radical(I*J); // compute radical of I*J_[1]=x2y+xy2> radical(intersect(I,J)); // compute radical of intersection_[1]=x2y+xy2> reduce(y3,std(I+J)); // gives 0 if and only if y^3 in I+J0

In this example it turned out that√

I · J =√

I∩ J. In fact, this is not a coincidence — the followinglemma and exercise show that the product and the intersection of ideals are very closely related.

Lemma 1.7 (Product and intersection of ideals). For any two ideals I and J in a ring R we have

1. Ideals 11

(a) I · J ⊂ I∩ J;

(b)√

I · J =√

I∩ J =√

I∩√

J.

Proof.

(a) It suffices to check that all generators of I ·J lie in I∩J. But for a ∈ I and b ∈ J it is obviousthat ab ∈ I∩ J, so the result follows.

(b) We show a circular inclusion, with√

I · J ⊂√

I∩ J following from (a).

If a ∈√

I∩ J then an ∈ I∩ J for some n ∈ N, so an ∈ I and an ∈ J, and hence a ∈√

I∩√

J.Finally, if a ∈

√I∩√

J then am ∈ I and an ∈ J for some m,n ∈ N, therefore am+n ∈ I · J andthus a ∈

√I · J. �

Exercise 1.8. Let I1, . . . , In be pairwise coprime ideals in a ring R. Prove that I1 · · · · ·In = I1∩·· ·∩In.

Exercise 1.9. Show that the ideal (x1, . . . ,xn)EK[x1, . . . ,xn] cannot be generated by fewer than nelements. Hence in particular, the polynomial ring K[x1, . . . ,xn] is not a principal ideal domain forn≥ 2.

We will see however in Remark 8.6 that these polynomial rings still admit unique prime factoriza-tions of its elements, so that the results of Example 1.4 continue to hold for principal ideals in theserings.

Remark 1.10 (Ideals of subvarieties = radical ideals). Radical ideals play an important role ingeometry: if Y is a subvariety of a variety X and f ∈ A(X) with f n ∈ I(Y ), then ( f (x))n = 0 forall x ∈ Y — but this obviously implies f (x) = 0 for all x ∈ Y , and hence f ∈ I(Y ). So ideals ofsubvarieties are always radical.

In fact, if the ground field K is algebraically closed, i. e. if every non-constant polynomial over Khas a zero (as e. g. for K = C), we will see in Corollary 10.14 that it is exactly the radical ideals inA(X) that are ideals of subvarieties. So in this case there is a one-to-one correspondence

{subvarieties of X} 1:1←→ {radical ideals in A(X)}

Y 7−→ I(Y )V (I) ←−7 I.

In other words, we have V (I(Y )) = Y for every subvariety Y of X (which we have already seen inLemma 0.9 (c)), and I(V (I)) = I for every radical ideal IEA(X). In order to simplify our geometricinterpretations we will therefore usually assume from now on in our geometric examples that theground field is algebraically closed and the above one-to-one correspondence holds. Note that thiswill not lead to circular reasoning as we will never use these geometric examples to prove anything.

Exercise 1.11.(a) Give a rigorous proof of the one-to-one correspondence of Remark 1.10 in the case of the

ambient variety A1C, i. e. between subvarieties of A1

C and radical ideals in A(A1C) = C[x].

(b) Show that this one-to-one correspondence does not hold in the case of the ground field R,i. e. between subvarieties of A1

R and radical ideals in A(A1R) = R[x].

Remark 1.12 (Geometric interpretation of operations on ideals). Let X be a variety over an alge-braically closed field, and let A(X) be its coordinate ring. Assuming the one-to-one correspondenceof Remark 1.10 between subvarieties of X and radical ideals in A(X) we can now give a geometricinterpretation of the operations of Construction 1.1:

(a) As I + J is the ideal generated by I∪ J, we have for any two (radical) ideals I,JEA(X)

V (I + J) = {x ∈ X : f (x) = 0 for all f ∈ I∪ J}= {x ∈ X : f (x) = 0 for all f ∈ I}∩{x ∈ X : f (x) = 0 for all f ∈ J}=V (I)∩V (J).

12 Andreas Gathmann

So the intersection of subvarieties corresponds to the sum of ideals. (Note however that thesum of two radical ideals may not be radical, so strictly speaking the algebraic operationcorresponding to the intersection of subvarieties is taking the sum of the ideals and then itsradical.)

Moreover, as the whole space X and the empty set /0 obviously correspond to the zero ideal(0) resp. the whole ring (1) = A(X), the condition I + J = A(X) that I and J are coprimetranslates into the intersection of V (I) and V (J) being empty.

(b) For any two subvarieties Y,Z of X

I(Y ∪Z) = { f ∈ A(X) : f (x) = 0 for all x ∈ Y ∪Z}= { f ∈ A(X) : f (x) = 0 for all x ∈ Y}∩{ f ∈ A(X) : f (x) = 0 for all x ∈ Z}= I(Y )∩ I(Z),

and thus the union of subvarieties corresponds to the intersection of ideals. As the productof ideals has the same radical as the intersection by Lemma 1.7 (b), the union of subvarietiesalso corresponds to taking the product of the ideals (and then its radical).

(c) Again for two subvarieties Y,Z of X we have

I(Y\Z) = { f ∈ A(X) : f (x) = 0 for all x ∈ Y\Z}= { f ∈ A(X) : f (x) ·g(x) = 0 for all x ∈ Y and g ∈ I(Z)}= { f ∈ A(X) : f · I(Z)⊂ I(Y )}= I(Y ) : I(Z),

so taking the set-theoretic difference Y\Z corresponds to quotient ideals. (Strictly speaking,the difference Y\Z is in general not a variety, so the exact geometric operation correspondingto quotient ideals is taking the smallest subvariety containing Y\Z.)

Summarizing, we obtain the following translation between geometric and algebraic terms:

SUBVARIETIES ←→ IDEALSfull space (0)empty set (1)intersection sumunion product / intersectiondifference quotientdisjoint subvarieties coprime ideals

Exercise 1.13. Show that the equation of ideals

(x3− x2,x2y− x2,xy− y,y2− y) = (x2,y)∩ (x−1,y−1)

holds in the polynomial ring C[x,y]. Is this a radical ideal? What is its zero locus in A2C?

As an example that links the concepts introduced so far, let us now consider the Chinese RemainderTheorem that you already know for the integers [G1, Proposition 11.22] and generalize it to arbitraryrings.

Proposition 1.14 (Chinese Remainder Theorem). Let I1, . . . , In be ideals in a ring R, and considerthe ring homomorphism

ϕ : R→ R/I1×·· ·×R/In, a 7→ (a, . . . ,a).

(a) ϕ is injective if and only if I1∩·· ·∩ In = (0).

(b) ϕ is surjective if and only if I1, . . . , In are pairwise coprime.

1. Ideals 13

Proof.

(a) This follows immediately from kerϕ = I1∩·· ·∩ In.

(b) “⇒” If ϕ is surjective then (1,0, . . . ,0) ∈ imϕ . In particular, there is an element a ∈ Rwith a = 1 mod I1 and a = 0 mod I2. But then 1 = (1− a)+ a ∈ I1 + I2, and henceI1 + I2 = R. In the same way we see Ii + I j = R for all i 6= j.

“⇐” Let Ii + I j = R for all i 6= j. In particular, for i = 2, . . . ,n there are ai ∈ I1 and bi ∈Ii with ai + bi = 1, so that bi = 1− ai = 1 mod I1 and bi = 0 mod Ii. If we thenset b := b2 · · · · · bn we get b = 1 mod I1 and b = 0 mod Ii for all i = 2, . . . ,n. So(1,0, . . . ,0) = ϕ(b) ∈ imϕ . In the same way we see that the other unit generators arein the image of ϕ , and hence ϕ is surjective. �

Example 1.15.(a) Consider the case R = Z, and let a1, . . . ,an ∈ Z be pairwise coprime. Then the residue class

mapϕ : Z→ Za1 ×·· ·×Zan , x 7→ (x, . . . ,x)

is surjective by Proposition 1.14 (b). Its kernel is (a1)∩·· ·∩(an) = (a) with a := a1 · · · · ·anby Exercise 1.8, and so by the homomorphism theorem [G1, Proposition 8.12] we obtain anisomorphism

Za→ Za1 ×·· ·×Zan , x 7→ (x, . . . ,x),

which is the well-known form of the Chinese Remainder Theorem for the integers [G1,Proposition 11.22].

(b) Let X be a variety, and let Y1, . . . ,Yn be subvarieties of X . Recall from Remark 0.13 that fori = 1, . . . ,n we have isomorphisms A(X)/I(Yi)∼= A(Yi) by restricting functions from X to Yi.Using the translations from Remark 1.12, Proposition 1.14 therefore states that the combinedrestriction map ϕ : A(X)→ A(Y1)×·· ·×A(Yn) to all given subvarieties is . . .

• injective if and only if the subvarieties Y1, . . . ,Yn cover all of X ;

• surjective if and only if the subvarieties Y1, . . . ,Yn are disjoint.

In particular, if X is the disjoint union of the subvarieties Y1, . . . ,Yn, then the Chinese Re-mainder Theorem says that ϕ is an isomorphism, i. e. that giving a function on X is the sameas giving a function on each of the subvarieties — which seems obvious from geometry.

In our study of ideals, let us now consider their behavior under ring homomorphisms.

Definition 1.16 (Contraction and extension). Let ϕ : R→ R′ be a ring homomorphism.

(a) For any ideal IER′ the inverse image ϕ−1(I) is an ideal of R. We call it the inverse imageideal or contraction of I by ϕ , sometimes denoted Ic if it is clear from the context whichmorphism we consider.

(b) For IER the ideal generated by the image ϕ(I) is called the image ideal or extension of Iby ϕ . It is written as ϕ(I) ·R′, or Ie if the morphism is clear from the context.

Remark 1.17.(a) Note that for the construction of the image ideal of an ideal IER under a ring homomorphism

ϕ : R→ R′ we have to take the ideal generated by ϕ(I), since ϕ(I) itself is in general not yetan ideal: take e. g. ϕ : Z→ Z[x] to be the inclusion and I = Z. But if ϕ is surjective thenϕ(I) is already an ideal and thus Ie = ϕ(I):

• for b1,b2 ∈ ϕ(I) we have a1,a2 ∈ I with b1 = ϕ(a1) and b2 = ϕ(a2), and so b1 +b2 =ϕ(a1 +a2) ∈ ϕ(I);

• for b ∈ ϕ(I) and s ∈ R′ we have a ∈ I and r ∈ R with ϕ(a) = b and ϕ(r) = s, and thussb = ϕ(ra) ∈ ϕ(I).

14 Andreas Gathmann

(b) If R is a field and R′ 6= {0} then any ring homomorphism ϕ : R→ R′ is injective: its kernelis 0 since an element a ∈ R\{0} with ϕ(a) = 0 would lead to the contradiction

1 = ϕ(1) = ϕ(a−1 a) = ϕ(a−1) ·ϕ(a) = 0.

This is the origin of the names “contraction” and “extension”, since in this case these twooperations really make the ideal “smaller” and “bigger”, respectively.

Remark 1.18 (Geometric interpretation of contraction and extension). As in Construction 0.11, letf : X → Y be a morphism of varieties, and let ϕ : A(Y )→ A(X), g 7→ g ◦ f be the associated mapbetween the coordinate rings.

(a) For any subvariety Z ⊂ X we have

I( f (Z)) = {g ∈ A(Y ) : g( f (x)) = 0 for all x ∈ Z}= {g ∈ A(Y ) : ϕ(g) ∈ I(Z)}

= ϕ−1(I(Z)),

so taking images of varieties corresponds to the contraction of ideals.

(b) For a subvariety Z ⊂ Y the zero locus of the extension I(Z)e by ϕ is

V (ϕ(I(Z))) = {x ∈ X : g( f (x)) = 0 for all g ∈ I(Z)}

= f−1({y ∈ Y : g(y) = 0 for all g ∈ I(Z)})

= f−1(V (I(Z)))

= f−1(Z)

by Lemma 0.9 (c). Hence, taking inverse images of subvarieties corresponds to the extensionof ideals.

So we can add the following two entries to our dictionary between geometry and algebra:

SUBVARIETIES ←→ IDEALSimage contractioninverse image extension

Exercise 1.19. Let ϕ : R→ R′ a ring homomorphism. Prove:

(a) I ⊂ (Ie)c for all IER;

(b) I ⊃ (Ic)e for all IER′;

(c) (IJ)e = Ie Je for all I,JER;

(d) (I∩ J)c = Ic∩ Jc for all I,JER′.

Exercise 1.20. Let f : X →Y be a morphism of varieties, and let Z and W be subvarieties of X . Thegeometric statements below are then obvious. Find and prove corresponding algebraic statementsfor ideals in rings.

(a) f (Z∪W ) = f (Z)∪ f (W );

(b) f (Z∩W )⊂ f (Z)∩ f (W );

(c) f (Z\W )⊃ f (Z)\ f (W ).

An important application of contraction and extension is that it allows an easy explicit descriptionof ideals in quotient rings.

1. Ideals 15

Lemma 1.21 (Ideals in quotient rings). Let I be an ideal in a ring R. Then contraction and extensionby the quotient map ϕ : R→ R/I give a one-to-one correspondence

{ideals in R/I} 1:1←→ {ideals J in R with J ⊃ I}

J 7−→ Jc

Je ←−7 J.

Proof. As the quotient map ϕ is surjective, we know by Remark 1.17 (a) that contraction and ex-tension are just the inverse image and image of an ideal, respectively. Moreover, it is clear that thecontraction of an ideal in R/I yields an ideal of R that contains I, and that the extension of an idealin R gives an ideal in R/I. So we just have to show that contraction and extension are inverse to eachother on the sets of ideals given in the lemma. But this is easy to check:

• For any ideal JER/I we have (Jc)e = ϕ(ϕ−1(J)) = J since ϕ is surjective.

• For any ideal JER with J ⊃ I we get

(Je)c = ϕ−1(ϕ(J)) = {a ∈ R : ϕ(a) ∈ ϕ(J)}= J+ I = J. �

Exercise 1.22. Let I ⊂ J be ideals in a ring R. By Lemma 1.21, the extension J/I of J by the quotientmap R→ R/I is an ideal in R/I. Prove that

(R/I)/(J/I)∼= R/J.02

At the end of this chapter, let us now consider ring homomorphisms from a slightly different pointof view that will also tell us which rings “come from geometry”, i. e. can be written as coordinaterings of varieties.

Definition 1.23 (Algebras and algebra homomorphisms). Let R be a ring.

(a) An R-algebra is a ring R′ together with a ring homomorphism ϕR′ : R→ R′.

(b) Let R1 and R2 be R-algebras with corresponding ring homomorphisms ϕR1 : R→R1 and ϕR2 :R→R2. A morphism or R-algebra homomorphism from R1 to R2 is a ring homomorphismϕ : R1→ R2 with ϕ ◦ϕR1 = ϕR2 .

It is often helpful to draw these maps in a diagram as shown on theright. Then the condition ϕ ◦ϕR1 = ϕR2 just states that this diagramcommutes, i. e. that any two ways along the arrows in the diagram hav-ing the same source and target — in this case the two ways to go fromR to R2 — will give the same map.

R

ϕR1 ϕR2

R1 R2ϕ

(c) Let R′ be an R-algebra with corresponding ring homomorphism ϕR′ : R → R′. An R-subalgebra of R′ is a subring R of R′ containing the image of ϕ . Note that R is then anR-algebra using the ring homomorphism ϕR : R→ R given by ϕR′ with the target restrictedto R. Moreover, the inclusion R→ R′ is an R-algebra homomorphism in the sense of (b).

In most of our applications, the ring homomorphism ϕR′ : R→ R′ needed to define an R-algebra R′

will be clear from the context, and we will write the R-algebra simply as R′. In fact, in many cases itwill even be injective. In this case we usually consider R as a subring of R′, drop the homomorphismϕR′ in the notation completely, and say that R ⊂ R′ is a ring extension. We will consider these ringextensions in detail in Chapter 9.

Remark 1.24. The ring homomorphism ϕR′ : R→ R′ associated to an R-algebra R′ can be used todefine a “scalar multiplication” of R on R′ by

R×R′→ R′, (a,c) 7→ a · c := ϕR′(a) · c.Note that by setting c = 1 this scalar multiplication determines ϕR′ back. So one can also think ofan R-algebra as a ring together with a scalar multiplication with elements of R that has the expectedcompatibility properties. In fact, one could also define R-algebras in this way.

16 Andreas Gathmann

Example 1.25.

(a) Without doubt the most important example of an algebra over a ring R is the polynomial ringR[x1, . . . ,xn], together with the obvious injective ring homomorphism R→ R[x1, . . . ,xn] thatembeds R into the polynomial ring as constant polynomials. In the same way, any quotientR[x1, . . . ,xn]/I of the polynomial ring by an ideal I is an R-algebra as well.

(b) Let X ⊂ AnK be a variety over a field K. Then its coordinate ring A(X) = K[x1, . . . ,xn]/I(X)

is a K-algebra by (a), with K mapping to A(X) as the constant functions. Moreover, the ringhomomorphism A(Y )→ A(X) of Construction 0.11 corresponding to a morphism f : X→Yto another variety Y is a K-algebra homomorphism, since composing a constant function withf gives again a constant function. In fact, one can show that these are precisely the mapsbetween the coordinate rings coming from morphisms of varieties, i. e. that Construction0.11 gives a one-to-one correspondence

{morphisms X → Y} 1:1←→ {K-algebra homomorphisms A(Y )→ A(X)}.

Definition 1.26 (Generated subalgebras). Let R′ be an R-algebra.

(a) For any subset M ⊂ R′ let

R[M] :=⋂

T⊃MR-subalgebra of R′

T

be the smallest R-subalgebra of R′ that contains M. We call it the R-subalgebra generatedby M. If M = {c1, . . . ,cn} is finite, we write R[M] = R[{c1, . . . ,cn}] also as R[c1, . . . ,cn].

(b) We say that R′ is a finitely generated R-algebra if there are finitely many c1, . . . ,cn withR[c1, . . . ,cn] = R′.

Remark 1.27. Note that the square bracket notation in Definition 1.26 is ambiguous: R[x1, . . . ,xn]can either mean the polynomial ring over R as in Definition 0.2 (if x1, . . . ,xn are formal variables),or the subalgebra of an R-algebra R′ generated by x1, . . . ,xn (if x1, . . . ,xn are elements of R′). Unfor-tunately, the usage of the notation R[x1, . . . ,xn] for both concepts is well-established in the literature,so we will adopt it here as well. Its origin lies in the following lemma, which shows that the elementsof an R-subalgebra generated by a set M are just the polynomial expressions in elements of M withcoefficients in R.

Lemma 1.28 (Explicit description of R[M]). Let M be a subset of an R-algebra R′. Then

R[M] =

{∑

i1,...,in∈Nai1,...,in ci1

1 · · · · · cinn : ai1,...,in ∈ R, c1, . . . ,cn ∈M, only finitely many ai1,...,in 6= 0

},

where multiplication in R′ with elements of R is defined as in Remark 1.24.

Proof. It is obvious that this set of polynomial expressions is an R-subalgebra of R′. Conversely,every R-subalgebra of R′ containing M must also contain these polynomial expressions, so the resultfollows. �

Example 1.29. In the field C of complex numbers the Z-algebra generated by the imaginary unit iis

Z[i] = { f (i) : f ∈ Z[x]}= {a+bi : a,b ∈ Z} ⊂ C

by Lemma 1.28. (Note again the double use of the square bracket notation: Z[x] is the polynomialring over Z, whereas Z[i] is the Z-subalgebra of C generated by i.)

Lemma 1.30 (Finitely generated R-algebras). An algebra R′ over a ring R is finitely generated ifand only if R′ ∼= R[x1, . . . ,xn]/I for some n ∈ N and an ideal I in the polynomial ring R[x1, . . . ,xn].

1. Ideals 17

Proof. Certainly, R[x1, . . . ,xn]/I is a finitely generated R-algebra since it is generated by the classesof x1, . . . ,xn. Conversely, let R′ be an R-algebra generated by c1, . . . ,cn ∈ S. Then

ϕ : R[x1, . . . ,xn]→ R′, ∑i1,...,in

ai1,...,in xi11 · · · · · x

inn 7→ ∑

i1,...,in

ai1,...,in ci11 · · · · · c

inn

is a ring homomorphism, and its image is precisely R[c1, . . . ,cn] = R′ by Lemma 1.28. So by thehomomorphism theorem [G1, Proposition 8.12] ϕ induces a ring isomorphism R[x1, . . . ,xn]/kerϕ ∼=R′, which by construction is also an R-algebra isomorphism. �

Remark 1.31 (Coordinate rings = reduced finitely generated K-algebras). Let K be an algebraicallyclosed field. Then by Remark 1.10 the coordinate rings of varieties over K are exactly the ringsof the form K[x1, . . . ,xn]/I for a radical ideal IEK[x1, . . . ,xn], so by Lemma 1.3 and Lemma 1.30precisely the reduced finitely generated K-algebras.

18 Andreas Gathmann

2. Prime and Maximal Ideals

There are two special kinds of ideals that are of particular importance, both algebraically and geo-metrically: the so-called prime and maximal ideals. Let us start by defining these concepts.

Definition 2.1 (Prime and maximal ideals). Let I be an ideal in a ring R with I 6= R.

(a) I is called a prime ideal if for all a,b ∈ R with ab ∈ I we have a ∈ I or b ∈ I. By induction,this is obviously the same as saying that for all a1, . . . ,an ∈ R with a1 · · · · ·an ∈ I one of theai must be in I.

(b) I is called a maximal ideal if there is no ideal J with I ( J ( R.

(c) The set of all prime ideals of R is called the spectrum, the set of all maximal ideals themaximal spectrum of R. We denote these sets by SpecR and mSpecR, respectively.

Remark 2.2. Let R 6= {0} be a ring. Note that its two trivial ideals R and (0) are treated differentlyin Definition 2.1:

(a) The whole ring R is by definition never a prime or maximal ideal. In fact, maximal idealsare just defined to be the inclusion-maximal ones among all ideals that are not equal to R.

(b) The zero ideal (0) may be prime or maximal if the corresponding conditions are satisfied.More precisely, by definition (0) is a prime ideal if and only if R is an integral domain [G1,Definition 7.6 (d)], and it is maximal if and only if there are no ideals except (0) and R, i. e.if R is a field [G1, Example 8.8 (c)].

In fact, there is a similar criterion for arbitrary ideals if one passes to quotient rings:

Lemma 2.3. Let I be an ideal in a ring R with I 6= R.

(a) I is a prime ideal if and only if R/I is an integral domain.

(b) I is a maximal ideal if and only if R/I is a field.

Proof.

(a) Passing to the quotient ring R/I, the condition of Definition 2.1 (a) says that I is prime if andonly if for all a,b ∈ R/I with a b = 0 we have a = 0 or b = 0, i. e. if and only if R/I is anintegral domain.

(b) By Lemma 1.21, the condition of Definition 2.1 (b) means exactly that the ring R/I has onlythe trivial ideals I/I and R/I, which is equivalent to R/I being a field [G1, Example 8.8(c)]. �

We can view this lemma as being analogous to Lemma 1.3, which asserted that I is a radical ideal ifand only if R/I is reduced. The fact that these properties of ideals are reflected in their quotient ringshas the immediate consequence that they are preserved under taking quotients as in Lemma 1.21:

Corollary 2.4. Let I ⊂ J be ideals in a ring R. Then J is radical / prime / maximal in R if and onlyif J/I is radical / prime / maximal in R/I.

Proof. By Lemma 1.3, the ideal J is radical in R if and only if R/J is reduced, and J/I is radical inR/I if and only if (R/I)/(J/I) is reduced. But these two rings are isomorphic by Exercise 1.22, sothe result follows.

The statement about prime and maximal ideals follows in the same way, using Lemma 2.3 insteadof Lemma 1.3. �

Corollary 2.5. Every maximal ideal in a ring is prime, and every prime ideal is radical.

2. Prime and Maximal Ideals 19

Proof. Passing to the quotient ring, this follows immediately from Lemma 1.3 and Lemma 2.3 sincea field is an integral domain and an integral domain is reduced. �

Example 2.6.(a) Let R be an integral domain, and let p ∈ R\{0} not be a unit, i. e. it does not have a multi-

plicative inverse in R [G1, Definition 7.6 (a)]. Then by definition the ideal (p) is prime ifand only if for all a,b ∈ R with p |ab we have p |a or p |b, i. e. by definition if and only if pis a prime element of R [G1, Definition 11.1 (b)]. Of course, this is the origin of the name“prime ideal”.

(b) We claim that for non-zero ideals in a principal ideal domain R the notions of prime andmaximal ideals agree. To see this, is suffices by Corollary 2.5 to show that every non-zeroprime ideal is maximal. So let IER be prime. Of course, we have I = (p) for some p ∈ R asR is a principal ideal domain, and since I 6= R by definition and I 6= 0 by assumption we knowby (a) that p is prime. Now if J ⊃ I is another ideal we must have J = (q) for some q | p. Butp is prime and thus irreducible, i. e. it cannot be written as a product of two non-units in R[G1, Definition 11.1 (a) and Lemma 11.3]. So up to a unit q must be 1 or p. But then J = Ror J = I, respectively, which means that I must have been maximal.

(c) Let K be a field, and consider the ideal I(a) = (x1−a1, . . . ,xn−an)EK[x1, . . . ,xn] of a for agiven point a = (a1, . . . ,an) ∈ An

K as in Example 0.7. Then the ring homomorphism

K[x1, . . . ,xn]/I(a)→ K, f 7→ f (a)

is obviously an isomorphism, since f ∈ I(a) is by definition equivalent to f (a) = 0. SoK[x1, . . . ,xn]/I(a)∼= K is a field, and thus by Lemma 2.3 (b) the ideal I(a) is maximal.

For general fields, not all maximal ideals of K[x1, . . . ,xn] have to be of this form. For exam-ple, the ideal (x2 +1)ER[x] is also maximal by (a) and (b), since the real polynomial x2 +1is irreducible and thus prime in R[x] [G1, Proposition 11.5]. But if K is algebraically closed,we will see in Corollary 10.10 that the ideals considered above are the only maximal idealsin the polynomial ring. In fact, it is easy to see that we would expect this if we look at thefollowing geometric interpretation of maximal ideals.

Remark 2.7 (Geometric interpretation of prime and maximal ideals). Let X be a variety over analgebraically closed field K, so that we have a one-to-one correspondence between subvarieties of Xand radical ideals in A(X) by Remark 1.10.

(a) As the correspondence between subvarieties and ideals reverses inclusions, the maximalideals of A(X) correspond to minimal subvarieties of X , i. e. to points of X . For example, wehave just seen in Example 2.6 (c) that the maximal ideal (x1−a1, . . . ,xn−an)EK[x1, . . . ,xn]is the ideal I(a) of the point a = (a1, . . . ,an) ∈ An

K .

(b) Let Y be a non-empty subvariety of X , corresponding to a proper ideal I(Y )EA(X).

If I(Y ) is not prime then there are functions f1, f2 ∈ A(X) such that f1 · f2 vanishes on Y , butf1 and f2 do not. Hence the zero loci Y1 :=VY ( f1) and Y2 :=VY ( f2) of f1 and f2 on Y are notall of Y , but their union is Y1∪Y2 =VY ( f1 f2) = Y . So we can write Y as a non-trivial unionof two subvarieties. If a variety has this property we call it reducible, otherwise irreducible.As shown in the picture below, the union Y =V (x1x2) of the two coordinate axes in A2

R is atypical example of a reducible variety, with f1 = x1 and f2 = x2 in the notation above.

= ∪

Y =V (x1x2) Y1 =V (x1) Y2 =V (x2)

20 Andreas Gathmann

Conversely, if Y is reducible with Y = Y1∪Y2 for two proper subvarieties Y1 and Y2, we canfind f1 ∈ I(Y1)\I(Y ) and f2 ∈ I(Y2)\I(Y ). Then f1 f2 ∈ A(X) vanishes on Y although f1 andf2 do not, and thus I(Y ) is not a prime ideal.

Summarizing, we get the following correspondence:

SUBVARIETIES ←→ IDEALSirreducible prime idealpoint maximal ideal

Exercise 2.8. Which of the following ideals are prime, which ones are maximal in Z[x]?

I = (5,x3 +2x+3) J = (4,x2 + x+1,x2 + x−1)

Exercise 2.9. Let ϕ : R→ S be a ring homomorphism, and let IES. Show that:

(a) If I is radical, then so is ϕ−1(I).

(b) If I is prime, then so is ϕ−1(I).

(c) If I is maximal, then ϕ−1(I) need not be maximal.

Exercise 2.10. Let R be a ring.

(a) Let I1, . . . , InER, and let PER be a prime ideal. If P ⊃ I1 ∩ ·· · ∩ In, prove that P ⊃ Ik forsome k = 1, . . . ,n.

(b) Let IER, and let P1, . . . ,PnER be prime ideals. If I ⊂ P1 ∪ ·· · ∪Pn, prove that I ⊂ Pk forsome k = 1, . . . ,n.

(c) Show that the statement of (b) still holds if P1 is not necessarily prime (but P2, . . . ,Pn stillare).

Can you give a geometric interpretation of these statements?

Exercise 2.11. Let R be the ring of all continuous real-valued functions on the unit interval [0,1].Similarly to Definition 0.3 (c), for any subset S of R we denote by

V (S) := {a ∈ [0,1] : f (a) = 0 for all f ∈ S} ⊂ [0,1]

the zero locus of S. Prove:

(a) For all a ∈ [0,1] the ideal Ia := { f ∈ R : f (a) = 0} is maximal.

(b) If f1, . . . , fn ∈ R with V ( f1, . . . , fn) = /0, then f 21 + · · ·+ f 2

n is invertible in R.

(c) For any ideal IER with I 6= R we have V (I) 6= /0.

(d) The assignment [0,1]→mSpecR, a 7→ Ia gives a one-to-one correspondence between pointsin the unit interval and maximal ideals of R (compare this to Remark 2.7 (a)).

Exercise 2.12. Let R be a ring such that for all a ∈ R there is a natural number n > 1 with an = a.

(a) Show that every prime ideal of R is maximal.

(b) Give an example of such a ring which is neither a field nor the zero ring.

We have now studied prime and maximal ideals in some detail, but left out one important question:whether such ideals actually always exist. More precisely, if I is an ideal in a ring R with I 6= R, is itguaranteed that there is a maximal (and hence also prime) ideal that contains I? In particular, takingI = (0), is it clear that there is any maximal ideal in R at all? From a geometric point of view thisseems to be trivial: using the translations of Remark 2.7 (a) we are just asking whether a non-emptyvariety always contains a point — which is of course true. But we know already that not every ringis the coordinate ring of a variety, so for general rings we have to find an algebraic argument thatensures the existence of maximal ideals. It turns out that this is rather tricky, so let us start by givingthe idea how such maximal ideals could be found.


Remark 2.13 (Naive strategy to find maximal ideals). Let I be an ideal in a ring R with I 6= R; wewant to find a maximal ideal that contains I. Set I0 := I. Of course, there is nothing to be done if I0is already maximal, so let us assume that it is not. Then there is an ideal I1 with I0 ( I1 ( R. Nowif I1 is maximal we are done, otherwise we find an ideal I2 with I0 ( I1 ( I2 ( R. Continuing thisprocess, we either find a maximal ideal containing I after a finite number of steps, or arrive at aninfinite strictly increasing chain

I0 ( I1 ( I2 ( · · ·of ideals in R.

Is it possible that such an infinite chain of ideals exists? In general, the answer to this question isyes (see e. g. Example 7.2 (c)). There is a special and very important class of rings however — theNoetherian rings that we will discuss in Chapter 7 — that do not admit such infinite increasing chainsof ideals. In these rings the above process must terminate, and so we can easily find a maximal idealcontaining I. In fact, all coordinate rings of varieties turn out to be Noetherian (see Remark 7.15),which is why our geometric picture above suggested that the existence of maximal ideals might betrivial. 03

But what can we do if the chain above does not terminate? The main observation is that we can thenform a “limit”

I∞ :=⋃

n∈NIn

over all previous ideals (it is easy to see that this is in fact an ideal which is not equal to R; wewill check this in the proof of Corollary 2.17). Of course, I∞ is strictly bigger than every In. So itis an ideal that we have not seen before in the chain, and consequently we can continue the aboveprocess with this limit ideal: if it is maximal we are finished, otherwise choose a bigger one whichwe might call I∞+1, and so on. Unless we find a maximal ideal at some point, we can thus constructanother infinite increasing chain starting with I∞, take the limit over this chain again, repeat thewhole process of constructing a chain and taking its limit infinitely many times, take the limit overall these infinitely many limit ideals, and so on. Intuitively speaking, we obtain an increasing “chain”of ideals in this way that is really long, certainly not indexed any more by the natural numbers, andnot even countable. Continuing our sloppy wording, we could hope that eventually we would haveto obtain even more ideals in this way than there are subsets of R at all. This would of course be acontradiction, suggesting that the process must stop at some point and give us a maximal ideal.

In fact, we will now make this argument precise. The key step in the proof is Zorn’s Lemma,which abstracts from the above situation of rings and ideals and roughly states that in any sort ofstructure where you can compare objects and the above “limiting process over chains” works to getsomething bigger than what you had before, you will find a maximal object. To be able to formulatethis rigorously we need some definitions regarding orders on sets.

Definition 2.14 (Orders). Let ≤ be a relation on a set M.

(a) We call ≤ a partial order on M if:

(i) a≤ a for all a ∈M (we say ≤ is reflexive);

(ii) for all a,b,c ∈M with a≤ b and b≤ c we have a≤ c (we say ≤ is transitive);

(iii) for all a,b ∈M with a≤ b and b≤ a we have a = b (we say ≤ is antisymmetric).

(b) If in addition a≤ b or b≤ a for all a,b ∈M, then ≤ is called a total order on M.

(c) An element b ∈M is an upper bound for a subset A⊂M if a≤ b for all a ∈ A.

(d) An element a ∈M is called maximal if for all b ∈M with a≤ b we have b = a.

For a partial order ≤ we write a < b if a≤ b and a 6= b. A set M together with a partial or total orderis called a partially or totally ordered set, respectively.

Example 2.15.(a) The sets N, Z, Q, and R with their standard order are all totally ordered sets.

22 Andreas Gathmann

(b) Certainly the most important example of a partial order (in fact, probably our only example)is the set-theoretic inclusion, where M is any family of sets. Note that this is in general nota total order since for two sets A,B ∈ M it might of course happen that neither A ⊂ B norB⊂ A holds. But if we have a chain A0 ( A1 ( A2 ( · · · of sets (or even a “longer chain” asin Remark 2.13) then the family {A0,A1,A2, . . .} of all sets in this chain is totally ordered byinclusion. So totally ordered sets will be our generalization of chains to families that are notnecessarily indexed by the natural numbers.

(c) If M is the set of all ideals I 6= R in a ring R, partially ordered by inclusion as in (b), then themaximal elements of M are by definition exactly the maximal ideals of R.

(d) In contrast to the case of total orders, a maximal element a in a partially ordered set M neednot be an upper bound for M, because a might not be comparable to some of the elements ofM.

With this notation we can thus translate our informal condition in Remark 2.13 that “the limitingprocess works” into the rigorous requirement that every totally ordered set should have an upperbound. If this condition is met, Zorn’s Lemma guarantees the existence of a maximal element andthus makes the intuitive argument of Remark 2.13 precise:

Proposition 2.16 (Zorn’s Lemma). Let M be a partially ordered set in which every totally orderedsubset has an upper bound. Then M has a maximal element.

Before we prove Zorn’s Lemma, let us first apply it to the case of ideals in rings to see that it reallysolves our existence problem for maximal ideals.

Corollary 2.17 (Existence of maximal ideals). Let I be an ideal in a ring R with I 6= R. Then I iscontained in a maximal ideal of R.

In particular, every ring R 6= 0 has a maximal ideal.

Proof. Let M be the set of all ideals JER with J ⊃ I and J 6= R. By Example 2.15 (c), the maximalideals of R containing I are exactly the maximal elements of M, and hence by Zorn’s Lemma itsuffices to show that every totally ordered subset of M has an upper bound.

So let A ⊂M be a totally ordered subset, i. e. a family of proper ideals of R containing I such that,for any two of these ideals, one is contained in the other. If A = /0 then we can just take I ∈M as anupper bound for A. Otherwise, let

J′ :=⋃J∈A

J

be the union of all ideals in A. We claim that this is an ideal:

• 0 ∈ J′, since 0 is contained in each J ∈ A, and A is non-empty.

• If a1,a2 ∈ J′, then a1 ∈ J1 and a2 ∈ J2 for some J1,J2 ∈ A. But A is totally ordered, so withoutloss of generality we can assume that J1 ⊂ J2. It follows that a1 +a2 ∈ J2 ⊂ J′.

• If a ∈ J′, i. e. a ∈ J for some J ∈ A, then ra ∈ J ⊂ J′ for any r ∈ R.

Moreover, J′ certainly contains I, and we must have J′ 6= R since 1 /∈ J for all J ∈ A, so that 1 /∈ J′.Hence J′ ∈M, and it is certainly an upper bound for A. Thus, by Zorn’s Lemma, M has a maximalelement, i. e. there is a maximal ideal in R containing I. �

So to complete our argument we have to give a proof of Zorn’s Lemma. However, as most textbooksusing Zorn’s Lemma do not prove it but rather say that it is simply an axiom of set theory, let us firstexplain shortly in what sense we can prove it.

Remark 2.18 (Zorn’s Lemma and the Axiom of Choice). As you know, essentially all of mathemat-ics is built up on the notion of sets. Nevertheless, if you remember your first days at university, youwere not given precise definitions of what sets actually are and what sort of operations you can do


with them. One usually just uses the informal statement that a set is a “collection of distinct objects”and applies common sense when dealing with them.

Although this approach is good to get you started, it is certainly not satisfactory from a mathemat-ically rigorous point of view. In fact, it is even easy to obtain contradictions (such as Russell’sParadox [G2, Remark 1.14]) if one applies common sense too naively when working with sets. Soone needs strict axioms for set theory — the ones used today were set up by Zermelo and Fraenkelaround 1930 — that state exactly which operations on sets are allowed. We do not want to list allthese axioms here, but as a first approximation one can say that one can always construct new setsfrom old ones, whereas “circular definitions” (that try e. g. to construct a set that contains itself as anelement) are forbidden.

Of course, the idea of these axioms is that they are all “intuitively obvious”, so that nobody willhave problems to accept them as the foundation for all of mathematics. One of them is the so-calledAxiom of Choice; it states that if you have a collection of non-empty sets you can simultaneouslychoose an element from each of them (even if you do not have a specific rule to make your choice).For example, if you want to prove that a surjective map f : A→ B has a right-sided inverse, i. e. amap g : B→ A with f ◦g = idB, you need to apply the Axiom of Choice since you have to constructg by simultaneously choosing an inverse image of every element of B under f . In a similar way youhave probably used the Axiom of Choice many times already without knowing about it — simplybecause it seems intuitively obvious.

Now it happens that Zorn’s Lemma is in fact equivalent to the Axiom of Choice. In other words,if we removed the Axiom of Choice from the axioms of set theory we could actually prove it if wetook Zorn’s Lemma as an axiom instead. But nobody would want to do this since the statement ofthe Axiom of Choice is intuitively clear, whereas Zorn’s Lemma is certainly not. So it seems a bitcheated not to prove Zorn’s Lemma because it could be taken as an axiom.

Having said all this, what we want to do now is to assume the Axiom of Choice (which you havedone in all your mathematical life anyway) and to prove Zorn’s Lemma with it. To do this, we needone more notion concerning orders.

Definition 2.19 (Well-ordered sets). A totally ordered set M is called well-ordered if every non-empty subset A of M has a minimum, i. e. an element a ∈ A such that a≤ b for all b ∈ A.

Example 2.20.(a) Any finite totally ordered set is well-ordered. Every subset of a well-ordered set is obviously

well-ordered, too.

(b) The set N of natural numbers is well-ordered with its standard order, whereas Z, Q, andR are not. Instead, to construct “bigger” well-ordered sets than N one has to add “infiniteelements”: the set N∪{∞} (obtained from N by adding one more element which is defined tobe bigger than all the others) is well-ordered. One can go on like this and obtain well-orderedsets N∪{∞,∞+1}, N∪{∞,∞+1,∞+2}, and so on.

It can be seen from these examples already that well-ordered sets are quite similar to the chains ofideals constructed in Remark 2.13. In fact, the idea of Remark 2.13 should be clearly visible in thefollowing proof of Zorn’s Lemma, in which our chains of ideals correspond to the f -sets introducedbelow, and the choice of a new bigger ideal that extends a previously constructed chain is given bythe function f .

Proof of Proposition 2.16 (Zorn’s Lemma). Let M be a partially ordered set in which every well-ordered subset has an upper bound (this is all we will need — so we could in fact weaken theassumption of Proposition 2.16 in this way). We will prove Zorn’s Lemma by contradiction, soassume that M has no maximal element.

For any well-ordered subset A ⊂M there is then an upper bound which cannot be maximal, and sowe can find an element of M which is even bigger, and thus strictly bigger than all elements of A.Choose such an element and call it f (A) — we can thus consider f as a function from the set of all

24 Andreas Gathmann

well-ordered subsets of M to M. (Actually, this is the point where we apply the Axiom of Choice asexplained in Remark 2.18.)

Let us call a subset A⊂M an f -set (we choose this name to indicatethat this notion depends on the choice of f made above) if it is well-ordered and satisfies a = f (A<a) for all a ∈ A, where we used theobvious notation A<a := {b ∈ A : b < a}. Drawing A symbolicallyas a line (since it is a totally ordered set after all), one can visualizethis condition as in the picture on the right.

A

A<a

a

f

Intuitively, an f -set is thus determined at each point a ∈ A by its predecessors in A by applying f .For example:

• If an f -set A has only finitely many elements a1 < · · · < an, we must have a1 = f ( /0) andai = f ({a1, . . . ,ai−1}) for i = 2, . . . ,n.

• If A is an f -set, then A∪{ f (A)} is also an f -set, obtained by “adding one element at theend” — and this is in fact the only element we could add at the end to obtain a new f -set.

In particular, we would expect that, although any two f -sets A and B might have different lengths,they should contain the same elements up to the point where one of them ends — so they shouldlook like the picture below on the left, and not like the one on the right. Let us prove this rigorously:

A Bb0

B

A

b0C

b

a

Claim: If A and B are two f -sets and there is an element b0 ∈ B\A, then A⊂ B and b0 is bigger thanall elements of A.

To prove this, let C be the union of all subsets of A∩B with the property that with any element theyalso contain all smaller elements in A∪B — let us call such subsets of A∩B saturated. Of course,C is then saturated as well, so it is obviously the biggest saturated set. We can think of it as the partwhere A and B still coincide, as in the picture above.

By construction, it is clear that C ⊂ A. If we had C 6= A, then A\C and B\C would be non-empty(the latter because it contains b0), and so there are a = min(A\C) and b = min(B\C) since A and Bare well-ordered. Then A<a = B<b =C by construction, and so a = f (A<a) = f (B<b) = b as A andB are f -sets. But this means that C∪{a} is a bigger saturated set than C, which is a contradictionand shows that we must have C = A. So A =C is a subset of B, and b0 is bigger than all elements ofC = A, proving our claim.

Now let D be the union of all f -sets. Then every a ∈ D is contained in an f -set A, and by our claimall elements of D\A (which must be contained in another f -set) are bigger than a. Hence D is anf -set as well:

• D is totally ordered (an element a of an f -set A is smaller than all elements of D\A, and canbe compared to all other elements of A since A is totally ordered);

• a minimum of any non-empty subset D′ ⊂ D can be found in any f -set A with A∩D′ 6= /0,since the other elements of D′ are bigger anyway — so D is well-ordered;

• for any a ∈ D in an f -set A we have f (D<a) = f (A<a) = a.

So D is the biggest f -set of M. But D∪{ f (D)} is an even bigger f -set, which is a contradiction.Hence M must have a maximal element. �

As another application of Zorn’s lemma, let us prove a formula for the radical of an ideal in terms ofprime ideals.


Lemma 2.21. For every ideal I in a ring R we have√

I =⋂

P primeP⊃I

P.

Proof.

“⊂” If a ∈√

I then an ∈ I for some n. But then also an ∈ P for every prime ideal P ⊃ I, whichimplies a ∈ P by Definition 2.1 (a).

“⊃” Let a ∈ R with a /∈√

I, i. e. an /∈ I for all n ∈ N. Consider the set

M = {J : JER with J ⊃ I and an /∈ J for all n ∈ N}.

In the same way as in the proof of Corollary 2.17 we see that every totally ordered subsetof M has an upper bound (namely I if the subset is empty, and the union of all ideals in thesubset otherwise). Hence by Proposition 2.16 there is a maximal element P of M. It sufficesto prove that P is prime, for then we have found a prime ideal P ⊃ I with a /∈ P, so that adoes not lie in the right hand side of the equation of the lemma.

So assume that we have b,c ∈ R with bc ∈ P, but b /∈ P and c /∈ P. Then P+(b) and P+(c)are strictly bigger than P, and thus by maximality cannot lie in M. This means that there aren,m ∈ N such that an ∈ P+(b) and am ∈ P+(c), from which we obtain

an+m ∈ (P+(b)) · (P+(c))⊂ P+(bc) = P,

in contradiction to P ∈M. Hence P must be prime, which proves the lemma. �

Remark 2.22. Let Y be a subvariety of a variety X . Then the statement of Lemma 2.21 for the(already radical) ideal I(Y ) in the ring A(X) corresponds to the geometrically obvious statement thatthe variety Y is the union of its irreducible subvarieties (see Remark 2.7).

Exercise 2.23 (Minimal primes). Let I be an ideal in a ring R with I 6= R. We say that a prime idealPER is minimal over I if I ⊂ P and there is no prime ideal Q with I ⊂ Q ( P.

(a) Prove that there is always a minimal prime ideal over I.

(b) Determine all minimal prime ideals over (x2y,xy2) in R[x,y].

If R is the coordinate ring of a variety and I the ideal of a subvariety, what is the geometric interpre-tation of a prime ideal minimal over I?

Exercise 2.24. Let P and Q be two distinct maximal ideals in a ring R, and let n,m ∈ N. Show thatPn and Qm are coprime.

Exercise 2.25. Show that for any ring R the following three statements are equivalent:

(a) R has exactly one prime ideal.

(b) Every element of R is either a unit or nilpotent.

(c)√(0) is a maximal ideal.

Give an example for such a ring which is not a field.

Exercise 2.26. Let M be an infinite set.

(a) Use Zorn’s Lemma to prove that M can be written as a disjoint union of sets that are allcountably infinite.

(b) Show that for any non-empty and at most countable set A there is a bijection between M andM×A.

Exercise 2.27.(a) Show that every vector space has a basis (even if it is not finitely generated).

26 Andreas Gathmann

(b) Let n,m ∈ N>0 with n 6= m. Of course, you know that Rn and Rm are not isomorphic asR-vector spaces. However, prove now that Rn and Rm are isomorphic as groups (with thestandard addition).

(Hint: Consider Rn and Rm as Q-vector spaces, and use Exercise 2.26 (b).)04

3. Modules 27

3. Modules

In linear algebra, the most important structure is that of a vector space over a field. For commutativealgebra it is therefore useful to consider the generalization of this concept to the case where theunderlying space of scalars is a commutative ring R instead of a field. The resulting structure iscalled a module; we will introduce and study it in this chapter.

In fact, there is another more subtle reason why modules are very powerful: they unify many otherstructures that you already know. For example, when you first heard about quotient rings you wereprobably surprised that in order to obtain a quotient ring R/I one needs an ideal I of R, i. e. astructure somewhat different from that of a (sub-)ring. In contrast, we will see in Example 3.4 (a)that ideals as well as quotient rings of R are just special cases of modules over R, so that one can dealwith both these structures in the same way. Even more unexpectedly, it turns out that modules overcertain rings allow a special interpretation: modules over Z are nothing but Abelian groups, whereasa module over the polynomial ring K[x] over a field K is exactly the same as a K-vector space Vtogether with a linear map ϕ : V → V (see Examples 3.2 (d) and 3.8, respectively). Consequently,general results on modules will have numerous consequences in many different setups.

So let us now start with the definition of modules. In principle, their theory that we will thenquickly discuss in this chapter is entirely analogous to that of vector spaces [G2, Chapters 13 to 18].However, although many properties just carry over without change, others will turn out to be vastlydifferent. Of course, proofs that are literally the same as for vector spaces will not be repeated here;instead we will just give references to the corresponding well-known linear algebra statements inthese cases.

Definition 3.1 (Modules). Let R be a ring. An R-module is a set M together with two operations

+ : M×M→M and · : R×M→M

(an “addition” in M and a “scalar multiplication” with elements of R) such that for all m,n ∈M anda,b ∈ R we have:

(a) (M,+) is an Abelian group;

(b) (a+b) ·m = a ·m+b ·m and a · (m+n) = a ·m+a ·n;

(c) (a ·b) ·m = a · (b ·m);

(d) 1 ·m = m.

We will also call M a module over R, or just a module if the base ring is clear from the context.

Example 3.2.(a) For a field R, an R-module is by definition exactly the same as an R-vector space [G2, Defi-

nition 13.1].

(b) Of course, the zero set {0} is a module, which we often simply write as 0.

(c) For n ∈N>0 the set Rn = {(a1, . . . ,an) : a1, . . . ,an ∈ R} is an R-module with componentwiseaddition and scalar multiplication. More generally, for two R-modules M and N the productM×N with componentwise addition and scalar multiplication is an R-module again.

(d) A Z-module is just the same as an Abelian group. In fact, any Z-module is an Abelian groupby definition 3.1 (a), and in any Abelian group (M,+) we can define a multiplication withintegers in the usual way by (−1) ·m :=−m and a ·m := m+ · · ·+m (a times) for a ∈N andm ∈M.

(e) Any R-algebra M is also an R-module by Remark 1.24, if we just forget about the possibilityto multiply two elements of M.

28 Andreas Gathmann

Definition 3.3 (Submodules, sums, and quotients). Let M be an R-module.

(a) A submodule of M is a non-empty subset N ⊂M satisfying m+ n ∈ N and am ∈ N for allm,n ∈ N and a ∈ R. We write this as N ≤M. Of course, N is then an R-module itself, withthe same addition and scalar multiplication as in M.

(b) For any subset S⊂M the set

〈S 〉 := {a1m1 + · · ·+anmn : n ∈ N,a1, . . . ,an ∈ R,m1, . . . ,mn ∈ S} ⊂M

of all R-linear combinations of elements of S is the smallest submodule of M that containsS. It is called the submodule generated by S. If S = {m1, . . . ,mn} is finite, we write 〈S 〉=〈{m1, . . . ,mn}〉 also as 〈m1, . . . ,mn 〉. The module M is called finitely generated if M = 〈S 〉for a finite set S⊂M.

(c) For submodules N1, . . . ,Nn ≤M their sum

N1 + · · ·+Nn = {m1 + · · ·+mn : mi ∈ Ni for all i = 1, . . . ,n}is obviously a submodule of M again. If moreover every element m ∈ N1 + · · ·+Nn has aunique representation as m = m1 + · · ·+mn with mi ∈ Ni for all i, we call N1 + · · ·+Nn adirect sum and write it also as N1⊕·· ·⊕Nn.

(d) If N ≤M is a submodule, the set

M/N := {x : x ∈M} with x := x+N

of equivalence classes modulo N is again a module [G2, Proposition 15.15], the so-calledquotient module of M modulo N.

Example 3.4.(a) Let R be a ring. If we consider R itself as an R-module, a submodule of R is by definition

the same as an ideal I of R. Moreover, the quotient ring R/I is then by Definition 3.3 (d) anR-module again.

Note that this is the first case where modules and vector spaces behave in a slightly differentway: if K is a field then the K-vector space K has no non-trivial subspaces.

(b) The polynomial ring K[x1, . . . ,xn] over a field K is finitely generated as a K-algebra (byx1, . . . ,xn), but not finitely generated as a K-module, i. e. as a K-vector space (the monomials1,x1,x2

1, . . . are linearly independent). So if we use the term “finitely generated” we alwayshave to make sure to specify whether we mean “finitely generated as an algebra” or “finitelygenerated as a module”, as these are two different concepts.

Exercise 3.5. Let N be a submodule of a module M over a ring R. Show:

(a) If N and M/N are finitely generated, then so is M.

(b) If M is finitely generated, then so is M/N.

(c) If M is finitely generated, N need not be finitely generated.

Definition 3.6 (Morphisms). Let M and N be R-modules.

(a) A morphism of R-modules (or R-module homomorphism, or R-linear map) from M to Nis a map ϕ : M→ N such that

ϕ(m+n) = ϕ(m)+ϕ(n) and ϕ(am) = aϕ(m)

for all m,n ∈ M and a ∈ R. The set of all such morphisms from M to N will be denotedHomR(M,N) or just Hom(M,N); it is an R-module again with pointwise addition and scalarmultiplication.

(b) A morphism ϕ : M → N of R-modules is called an isomorphism if it is bijective. In thiscase, the inverse map ϕ−1 : N →M is a morphism of R-modules again [G2, Lemma 13.25(a)]. We call M and N isomorphic (written M ∼= N) if there is an isomorphism betweenthem.

3. Modules 29

Example 3.7.(a) For any ideal I in a ring R, the quotient map ϕ : R→ R/I, a 7→ a is a surjective R-module

homomorphism.

(b) Let M and N be Abelian groups, considered as Z-modules as in Example 3.2 (d). Then aZ-module homomorphism ϕ : M→ N is the same as a homomorphism of Abelian groups,since ϕ(m+n) = ϕ(m)+ϕ(n) already implies ϕ(am) = aϕ(m) for all a ∈ Z.

(c) For any R-module M we have HomR(R,M)∼= M: the maps

M→ HomR(R,M), m 7→ (R→M,a 7→ am) and HomR(R,M)→M, ϕ 7→ ϕ(1)

are obviously R-module homomorphisms and inverse to each other. On the other hand,the module HomR(M,R) is in general not isomorphic to M: for the Z-module Z2 we haveHomZ(Z2,Z) = 0 by (b), as there are no non-trivial group homomorphisms from Z2 to Z.

(d) If N1, . . . ,Nn are submodules of an R-module M such that their sum N1⊕·· ·⊕Nn is direct,the morphism

N1×·· ·×Nn→ N1⊕·· ·⊕Nn, (m1, . . . ,mn) 7→ m1 + · · ·+mn

is bijective, and hence an isomorphism. One therefore often uses the notation N1⊕·· ·⊕Nnfor N1×·· ·×Nn also in the cases where N1, . . . ,Nn are R-modules that are not necessarilysubmodules of a given ambient module M.

Example 3.8 (Modules over polynomial rings). Let R be a ring. Then an R[x]-module M is the sameas an R-module M together with an R-module homomorphism ϕ : M→M:

“⇒” Let M be an R[x]-module. Of course, M is then also an R-module. Moreover, multiplicationwith x has to be R-linear, so ϕ : M→M,m 7→ x ·m is an R-module homomorphism.

“⇐” If M is an R-module and ϕ : M→M an R-module homomorphism we can give M the struc-ture of an R[x]-module by setting x ·m := ϕ(m), or more precisely by defining scalar multi-plication by ( n

∑i=0

aixi)·m :=

n

∑i=0

aiϕi(m),

where ϕ i denotes the i-fold composition of ϕ with itself, and ϕ0 := idM .

Remark 3.9 (Images and kernels of morphisms). Let ϕ : M→N be a homomorphism of R-modules.

(a) For any submodule M′ ≤M the image ϕ(M′) is a submodule of N [G2, Lemma 13.21 (a)].In particular, ϕ(M) is a submodule of N, called the image of ϕ .

(b) For any submodule N′ ≤ N the inverse image ϕ−1(N′) is a submodule of M [G2, Lemma13.21 (b)]. In particular, ϕ−1(0) is a submodule of M, called the kernel of ϕ .

Proposition 3.10 (Isomorphism theorems).(a) For any morphism ϕ : M→ N of R-modules there is an isomorphism

M/kerϕ → imϕ, m 7→ ϕ(m).

(b) For R-modules N′ ≤ N ≤M we have

(M/N′)/(N/N′) ∼= M/N.

(c) For two submodules N,N′ of an R-module M we have

(N +N′)/N′ ∼= N/(N∩N′).

Proof. The proofs of (a) and (b) are the same as in [G2, Proposition 15.22] and Exercise 1.22,respectively. For (c) note that N → (N +N′)/N′, m 7→ m is a surjective R-module homomorphismwith kernel N∩N′, so the statement follows from (a). �

Exercise 3.11. Let N be a proper submodule of an R-module M. Show that the following statementsare equivalent:

30 Andreas Gathmann

(a) There is no submodule P of M with N ( P ( M.

(b) The module M/N has only the trivial submodules 0 and M/N.

(c) M/N ∼= R/I for a maximal ideal IER.

The concepts so far were all entirely analogous to the case of vector spaces. There are a few con-structions however that are only useful for modules due to the existence of non-trivial ideals in thebase ring. Let us introduce them now.

Definition 3.12 (IM, module quotients, annihilators). Let M be an R-module.

(a) For an ideal IER we set

IM := 〈{am : a ∈ I,m ∈M}〉= {a1m1 + · · ·+anmn : n ∈ N,a1, . . . ,an ∈ I,m1, . . . ,mn ∈M}.

Note that IM is a submodule of M, and M/IM is an R/I-module in the obvious way.

(b) For two submodules N,N′ ≤M the module quotient (not to be confused with the quotientmodules of Definition 3.3 (d)) is defined to be

N′ :N := {a ∈ R : aN ⊂ N′} ER.

In particular, for N′ = 0 we obtain the so-called annihilator

annN := annR N := {a ∈ R : aN = 0} ER

of N. The same definition can also be applied to a single element m ∈M instead of a sub-module N: we then obtain the ideals

N′ :m := {a ∈ R : am ∈ N′} and ann m := {a ∈ R : am = 0}of R.

Example 3.13.(a) If M, N, and N′ are submodules of the R-module R, i. e. ideals of R by Example 3.4 (a), the

product IM and quotient N′ : N of Definition 3.12 are exactly the product and quotient ofideals as in Construction 1.1.

(b) If I is an ideal of a ring R then annR(R/I) = I.

Let us recall again the linear algebra of vector spaces over a field K. At the point where we arenow, i. e. after having studied subspaces and morphisms in general, one usually restricts to finitelygenerated vector spaces and shows that every such vector space V has a finite basis. This makes Visomorphic to Kn with n = dimK V ∈ N [G2, Proposition 14.22]. In other words, we can describevectors by their coordinates with respect to some basis, and linear maps by matrices — which arethen easy to deal with.

For a finitely generated module M over a ring R this strategy unfortunately breaks down. Ultimately,the reason for this is that the lack of a division in R means that a linear relation among generators ofM cannot necessarily be used to express one of them in terms of the others (so that it can be droppedfrom the set of generators). For example, the elements m = 2 and n = 3 in the Z-module Z satisfythe linear relation 3m−2n = 0, but neither is m an integer multiple of n, nor vice versa. So althoughZ= 〈m,n〉 and these two generators are linearly dependent over Z, neither m nor n alone generatesZ.

The consequence of this is that a finitely generated module M need not have a linearly independentset of generators. But this means that M is in general not isomorphic to Rn for some n ∈ N, and thusthere is no obvious well-defined notion of dimension. It is in fact easy to find examples for this: Z2as a Z-module is certainly not isomorphic to Zn for some n.

So essentially we have two choices if we want to continue to carry over our linear algebra results onfinitely generated vector spaces to finitely generated modules:

3. Modules 31

• restrict to R-modules that are of the form Rn for some n ∈ N; or

• go on with general finitely generated modules, taking care of the fact that generating sys-tems cannot be chosen to be independent, and thus that the coordinates with respect to suchsystems are no longer unique.

In the rest of this chapter, we will follow both strategies to some extent, and see what they lead to.Let us start by considering finitely generated modules that do admit a basis.

Definition 3.14 (Bases and free modules). Let M be a finitely generated R-module.

(a) We say that a family (m1, . . . ,mn) of elements of M is a basis of M if the R-module homo-morphism

Rn→M, (a1, . . . ,an) 7→ a1m1 + · · ·+anmn

is an isomorphism.

(b) If M has a basis, i. e. if it is isomorphic to Rn for some n, it is called a free R-module.

Example 3.15. If I is a non-trivial ideal in a ring R then R/I is never a free R-module: there can beno isomorphism

ϕ : Rn→ R/I, (a1, . . . ,an) 7→ a1m1 + · · ·+anmn

since in any case ϕ(0, . . . ,0) = ϕ(a,0, . . . ,0) for every a ∈ I.

Exercise 3.16. Let R be an integral domain. Prove that a non-zero ideal IER is a principal ideal ifand only if it is a free R-module.

Remark 3.17 (Linear algebra for free modules). Let M and N be finitely generated, free R modules.

(a) Any two bases of M have the same number of elements: assume that we have a basis with nelements, so that M ∼= Rn as R-modules. Choose a maximal ideal I of R by Corollary 2.17.Then R/I is a field by Lemma 2.3 (b), and M/IM is an R/I-vector space by Definition 3.12(a). Its dimension is

dimR/I M/IM = dimR/I Rn/IRn = dimR/I(R/I)n = n,

and so n is uniquely determined by M. We call n the rank rkM of M.

(b) In the same way as for vector spaces, we see that HomR(Rm,Rn) is isomorphic to the R-module Mat(n×m,R) of n×m-matrices over R [G2, Proposition 16.11]. Hence, afterchoosing bases for M and N we also have HomR(M,N) ∼= Mat(n×m,R) with m = rkMand n = rkN [G2, Proposition 16.23].

(c) An R-module homomorphism ϕ : M→ M is an isomorphism if and only if its matrix A ∈Mat(m×m,R) as in (b) is invertible, i. e. if and only if there is a matrix A−1 ∈Mat(m×m,R)such that A−1 A = AA−1 = E is the unit matrix. As expected, whether this is the case can bechecked with determinants as follows.

(d) For a square matrix A ∈ Mat(m×m,R) the determinant detA is defined in the usual way[G2, Proposition 18.12]. It has all the expected properties; in particular there is an adjointmatrix A# ∈Mat(m×m,R) such that A# A = AA# = detA ·E (namely the matrix with (i, j)-entry (−1)i+ j detA′j,i, where A′j,i is obtained from A by deleting row j and column i) [G2,Proposition 18.20 (a)]. With this we can see that A is invertible if and only if detA is a unitin R:

“⇒” If there is an inverse matrix A−1 then 1 = detE = det(A−1 A) = detA−1 ·detA, so detAis a unit in R.

“⇐” If detA is a unit, we see from the equation A# A = AA# = detA ·E that (detA)−1 ·A# isan inverse of A.

05

So all in all finitely generated, free modules behave very much in the same way as vector spaces.However, most modules occurring in practice will not be free — in fact, submodules and quotientmodules of free modules, as well as images and kernels of homomorphisms of free modules, will in

32 Andreas Gathmann

general not be free again. So let us now also find out what we can say about more general finitelygenerated modules.

First of all, the notion of dimension of a vector space, or rank of a free module as in Remark 3.17(a), is then no longer defined. The following notion of the length of a module can often be used tosubstitute this.

Definition 3.18 (Length of modules). Let M be an R-module.

(a) A composition series for M is a finite chain

0 = M0 ( M1 ( M2 ( · · ·( Mn = M

of submodules of M that cannot be refined, i. e. such that there is no submodule N of M withMi−1 ( N ( Mi for any i = 1, . . . ,n. (By Exercise 3.11, this is equivalent to Mi/Mi−1 havingno non-trivial submodules for all i, and to Mi/Mi−1 being isomorphic to R modulo somemaximal ideal for all i).

The number n above will be called the length of the series.

(b) If there is a composition series for M, the shortest length of such a series is called the lengthof M and denoted lR(M) (in fact, we will see in Exercise 3.19 (b) that then all compositionseries have this length). Otherwise, we set formally lR(M) = ∞.

If there is no risk of confusion about the base ring, we write lR(M) also as l(M).

Exercise 3.19. Let M be an R-module of finite length, i. e. an R-module that admits a compositionseries. Show that:

(a) If N < M is a proper submodule of M then l(N)< l(M).

(b) Every composition series for M has length l(M).

(c) Every chain 0 = M0 ( M1 ( M2 ( · · · ( Mn = M of submodules of M can be refined to acomposition series for M.

Example 3.20.(a) Let V be a vector space over a field K. If V has finite dimension n, there is a chain

0 =V0 (V1 ( · · ·(Vn =V

of subspaces of V with dimK Vi = i for all i. Obviously, this chain cannot be refined. Hence itis a composition series for V , and we conclude by Exercise 3.19 (b) that l(V ) = n = dimK V .

On the other hand, if V has infinite dimension, there are chains of subspaces of V of anylength. By Exercise 3.19 this is only possible if l(V ) = ∞.

So for vector spaces over a field, the length is just the same as the dimension.

(b) There is no statement analogous to (a) for free modules over a ring: already Z has infinitelength over Z, since there are chains

0 ( (2n)( (2n−1)( · · ·( (2)( Zof ideals in Z of any length.

(c) Certainly, a module M of finite length must be finitely generated: otherwise there wouldbe infinitely many elements (mi)i∈N of M such that all submodules Mi = 〈m1, . . . ,mi 〉 aredistinct. But then 0 = M0 ( M1 ( M2 ( · · · is an infinite chain of submodules, which byExercise 3.19 is impossible for modules of finite length.

On the other hand, a finitely generated module need not have finite length, as we have seen in(b). In fact, we will study the relation between the conditions of finite generation and finitelength in more detail in Chapter 7.

Exercise 3.21. What are the lengths of Z8 and Z12 as Z-modules? Can you generalize this statementto compute the length of any quotient ring R/I as an R-module, where I is an ideal in a principalideal domain R?

3. Modules 33

Let us now show that the length of modules satisfies the same relations as the dimension of vectorspaces when taking sums, intersections, quotients, or images and kernels [G2, Corollary 15.30,Proposition 15.16, and Corollary 15.25].

Proposition 3.22 (Additivity of the length of modules). For any submodule N of an R-submoduleM we have

l(N)+ l(M/N) = l(M).

Proof. Let us assume first that l(M)< ∞. By Exercise 3.19 (c), the chain 0≤ N ≤M can be refinedto a composition series

0 = N0 ( · · ·( Nn = N = M0 ( · · ·( Mm = M (∗)for M, where l(M) = n+m by Exercise 3.19 (b). Of course, the first part of this chain is then acomposition series for N, and so l(N) = n. Moreover, setting Pi := Mi/N for i = 1, . . . ,m we obtaina chain

0 = P0 ( · · ·( Pm = M/Nin which Pi/Pi−1 ∼= Mi/Mi−1 by Proposition 3.10 (b). As these modules have no non-trivial submod-ules, we see that the above chain of length m is a composition series for M/N, so that we get thedesired result l(N)+ l(M/N) = n+m = l(M).

Conversely, if l(N) and l(M/N) are finite, there are composition series

0 = N0 ( · · ·( Nn = N and 0 = P0 ( · · ·( Pm = M/N

for N and M/N, respectively. Setting Mi := q−1(Pi) with the quotient map q : M → M/N for alli = 1, . . . ,m, we have Mi/N = Pi. So as above, Mi/Mi−1 ∼= Pi/Pi−1 has no non-trivial submodules,and we obtain a composition series (∗) for M. This means that M has finite length as well, and theabove argument can be applied to prove the equation l(N)+ l(M/N) = l(M) again.

The only remaining case is that both sides of the equation of the proposition are infinite — but thenof course the statement is true as well. �

Corollary 3.23.(a) For any two submodules M1,M2 of an R-module M we have

l(M1 +M2)+ l(M1∩M2) = l(M1)+ l(M2).

(b) For any R-module homomorphism ϕ : M→ N we have

l(kerϕ)+ l(imϕ) = l(M).

Proof.

(a) By Proposition 3.10 (c) we have (M1 +M2)/M2 ∼= M1/(M1 ∩M2). Calling this module Q,we obtain by Proposition 3.22

l(M1 +M2) = l(M2)+ l(Q) and l(M1) = l(M1∩M2)+ l(Q).

So if l(M2) = ∞ then l(M1 +M2) = ∞, and the statement of the corollary holds. The same istrue if l(M1∩M2) = ∞ and thus l(M1) = ∞. Otherwise, we obtain

l(Q) = l(M1 +M2)− l(M2) = l(M1)− l(M1∩M2),

and hence the corollary holds in this case as well.

(b) This is just Proposition 3.22 applied to the homomorphism theorem M/kerϕ ∼= imϕ ofProposition 3.10 (a). �

Remark 3.24. An easy consequence of Corollary 3.23 (b) is that for a homomorphism ϕ : M→Mfrom a module of finite length to itself we have

ϕ injective ⇔ ϕ surjective ⇔ ϕ bijective

as in [G2, Corollary 15.26], since ϕ is injective if and only if l(kerϕ) = 0, and surjective if and onlyif l(imϕ) = l(M) (see Exercise 3.19 (a)).

34 Andreas Gathmann

What happens in this statement if we consider a module M that is only finitely generated, but notnecessarily of finite length (see Example 3.20 (c))? It is actually easy to see that in this case aninjective morphism ϕ : M → M need not be bijective: the map ϕ : Z→ Z, m 7→ 2m is a simplecounterexample. In view of this example it is probably surprising that the statement that a surjectivemap is also bijective still holds — this is what we want to show in Corollary 3.28 below. The mainingredient in the proof is the following generalization of the Cayley-Hamilton theorem from linearalgebra.

Proposition 3.25 (Cayley-Hamilton). Let M be a finitely generated R-module, I an ideal of R, andϕ : M→M an R-module homomorphism with ϕ(M) ⊂ IM. Then there is a monic polynomial (i. e.its leading coefficient is 1)

χ = xn +an−1xn−1 + · · ·+a0 ∈ R[x]

with a0, . . . ,an−1 ∈ I and

χ(ϕ) := ϕn +an−1ϕ

n−1 + · · ·+a0 id = 0 ∈ HomR(M,M),

where ϕ i denotes the i-fold composition of ϕ with itself.

Proof. Let m1, . . . ,mn be generators of M. By assumption we have ϕ(mi) ∈ IM for all i, and thusthere are ai, j ∈ I with

ϕ(mi) =n

∑j=1

ai, j m j for all i = 1, . . . ,n.

Considering M as an R[x]-module by setting x ·m := ϕ(m) for all m ∈M as in Example 3.8, we canrewrite this as

n

∑j=1

(xδi, j−ai, j)m j = 0 with δi, j :=

{1 if i = j,0 if i 6= j

for all i = 1, . . . ,n. Note that the left hand side of this equation, taken for all i, gives us an elementof Mn. If we multiply this from the left with the adjoint matrix of (xδi, j−ai, j)i, j ∈Mat(n×n,R[x])as in Remark 3.17 (d), we get

det((xδi,k−ai,k)i,k) ·m j = 0

for all j. So χ := det((xδi,k − ai,k)i,k) acts as the zero homomorphism on M, and expanding thedeterminant shows that the non-leading coefficients of this polynomial lie in fact in I. �

Remark 3.26. If R is a field and thus M a finitely generated vector space, we can only take I = R inProposition 3.25. In the proof, we can then choose m1, . . . ,mn to be a basis of M, so that (ai, j)i, j is thematrix of ϕ with respect to this basis, and χ is the characteristic polynomial of ϕ [G2, Definitions19.16 and Remark 19.21]. So in this case the statement of Proposition 3.25 is just the ordinaryCayley-Hamilton theorem for endomorphisms of finite-dimensional vector spaces [G2, Exercise20.24]. For general rings however, the generators m1, . . . ,mn are no longer independent, and so thereare several choices for the matrix (ai, j)i, j.

The following easy consequence of Proposition 3.25 is usually attributed to Nakayama. In fact, thereare many versions of Nakayama’s lemma in the literature (we will also meet some other closelyrelated statements in Exercise 6.16), but this one is probably one of the strongest. It concerns theconstruction IM of Definition 3.12 (a) for an ideal I in a ring R and an R-module M. More precisely,let us assume that M 6= 0 and IM = M. Of course, if R is a field this is only possible if I = R, i. e.if 1 ∈ I. If R is a general ring however, it does not necessarily follow that 1 ∈ I — but Nakayama’sLemma states that in the case of a finitely generated module M there is at least an element a ∈ I thatacts as the identity on M, i. e. such that am = m for all m ∈M.

Corollary 3.27 (Nakayama’s Lemma). Let M be a finitely generated R-module, and I an ideal ofR with IM = M. Then there is an element a ∈ I with am = m for all m ∈M.

3. Modules 35

Proof. As M = IM we can apply Proposition 3.25 to ϕ = id and our given ideal I to obtaina0, . . . ,an−1 ∈ I such that

idn+an−1 idn−1+ · · ·+a0 id = (1+an−1 + · · ·+a0) id = 0 ∈ HomR(M,M).

Setting a :=−an−1−·· ·−a0 ∈ I, this just means that (1−a)m = 0, i. e. am = m for all m ∈M. �

Corollary 3.28. If M is a finitely generated R-module, any surjective homomorphism ϕ : M→M isan isomorphism.

Proof. As in Example 3.8, consider M as an R[x]-module by setting x ·m := ϕ(m) for all m ∈ M.Then x ·M = ϕ(M) = M since ϕ is surjective, so we can apply Corollary 3.27 with I = (x) to obtaina polynomial f ∈ (x), i. e. a polynomial f = anxn+an−1xn−1+ · · ·+a1x without constant coefficient,such that

f ·m = anϕn(m)+an−1ϕ

n−1(m)+ · · ·+a1ϕ(m) = m for all m ∈M.

But this means that ϕ(m) = 0 implies m = 0, and so ϕ is injective. �

Exercise 3.29. For a prime number p ∈ N consider the subring R = { ab : a,b ∈ Z, p 6 | b} of Q, and

let M =Q as an R-module.

(a) Show that R has exactly one maximal ideal I. Which one?

(In fact, this will be obvious once we have studied localizations — see Example 6.6 andCorollary 6.10.)

(b) For the ideal of (a), prove that IM = M, but there is no a ∈ I with am = m for all m ∈M.

(c) Find a “small” set of generators for M as an R-module. Can you find a finite one? A minimalone?

36 Andreas Gathmann

4. Exact Sequences

In the last chapter we have studied many structures related to modules, such as submodules, quotientmodules, and module homomorphisms together with their images and kernels. We now want tointroduce a very useful piece of notation that can be used to deal with all these concepts in a unifiedway: the so-called exact sequences. In many cases they provide an easy and almost “graphical” wayto express and prove statements that are in principle elementary, but tedious to write down withoutthis language due to the sheer number of spaces, morphisms, or relations involved.

Definition 4.1 (Exact sequences). Let R be a ring, and let n ∈ N≥3. A sequence

M1ϕ1−→M2

ϕ2−→ ·· · ϕn−1−→Mn

of R-modules M1, . . . ,Mn and R-module homomorphisms ϕi : Mi→Mi+1 for i = 1, . . . ,n−1 is calledexact at position i∈ {2, . . . ,n−1} if imϕi−1 = kerϕi. It is called exact if it is exact at every positioni ∈ {2, . . . ,n−1}.In such sequences, we will often not label an arrow with its morphism if it is clear from the contextwhat the morphism is. In particular, this is the case if the source or target of the map is the zeromodule, so that the map is necessarily the zero morphism.

Example 4.2 (Exact sequences with few modules).

(a) A sequence 0−→Mϕ−→ N is exact if and only if kerϕ = 0, i. e. if and only if ϕ is injective.

A sequence Mϕ−→N −→ 0 is exact if and only if imϕ = N, i. e. if and only if ϕ is surjective.

(b) The sequence 0−→M −→ 0 is exact if and only if M = 0.

By (a), the sequence 0−→Mϕ−→N −→ 0 is exact if and only if ϕ is injective and surjective,

i. e. if and only if ϕ is an isomorphism.

Example 4.3 (Short exact sequences). By Example 4.2, the first interesting case of an exact sequenceoccurs if it has at least three non-zero terms. Therefore, an exact sequence of the form

0−→M1ϕ−→M2

ψ−→M3 −→ 0 (∗)is called a short exact sequence. There are two main sources for such short exact sequences:

(a) For any R-module homomorphism ψ : M→ N the sequence

0−→ kerψ −→Mψ−→ imψ −→ 0

is exact, where the first map is the inclusion of kerψ in M.

(b) For any submodule N of an R-module M the sequence

0−→ N −→M −→M/N −→ 0

is exact, where the first map is again the inclusion, and the second the quotient map.

In fact, up to isomorphisms every short exact sequence is of these forms: in a short exact sequence asin (∗) the second map ψ is surjective, and thus imψ = M3. Moreover, kerψ = imϕ = ϕ(M1)∼= M1,where the last isomorphism follows from the injectivity of ϕ . So the given sequence is of the typeas in (a). If we set N = kerψ ≤M2 and use the homomorphism theorem imψ ∼= M2/kerψ = M2/Nof Proposition 3.10 (a), we can also regard it as a sequence as in (b).

A nice feature of exact sequences is that there are simple rules to create new sequences from oldones. The simplest way to do this is to split and glue exact sequences as follows.

4. Exact Sequences 37

Lemma 4.4 (Splitting and gluing exact sequences).

(a) (Splitting) If

M1ϕ1−→M2

ϕ2−→M3ϕ3−→M4 (A)

is an exact sequence of R-modules, the two sequences

M1ϕ1−→M2

ϕ2−→ N −→ 0 and 0−→ N −→M3ϕ3−→M4 (B)

are also exact, where N = imϕ2 = kerϕ3, and the middle map in the second sequence is theinclusion of kerϕ3 in M3.

(b) (Gluing) Conversely, if

M1ϕ1−→M2

ϕ2−→ N −→ 0 and 0−→ N −→M3ϕ3−→M4 (B)

are two exact sequences, with N ≤ M3 a submodule and the middle map in the second se-quence the inclusion, the sequence

M1ϕ1−→M2

ϕ2−→M3ϕ3−→M4 (A)

is also exact.

Proof.

(a) The first sequence of (B) is exact at M2 since imϕ1 = kerϕ2, and exact at N as N = imϕ2.The second sequence of (B) is exact N as the middle map is an inclusion, and exact at M3since N = kerϕ3.

(b) The sequence of (A) is exact at M2 since imϕ1 = kerϕ2. Moreover, in (B) exactness of thefirst sequence at N and of the second sequence at M3 means that imϕ2 = N = kerϕ3, whichimplies that (A) is exact at M3. �

Remark 4.5 (Splitting an exact sequence into short ones). With Lemma 4.4 (a) every exact sequence

0−→M1ϕ1−→M2

ϕ2−→ ·· · ϕn−1−→Mn −→ 0 (∗)

(for simplicity with zero modules at the end) can be split up into short exact sequences

0−→ kerϕi −→Miϕi−→ imϕi −→ 0

for i = 2, . . . ,n− 1 as in Example 4.3 (a), where M1 = kerϕ2 and Mn = imϕn−1. Conversely, suchshort exact sequences with M1 = kerϕ2, Mn = imϕn−1, and imϕi−1 = kerϕi for i = 3, . . . ,n−1 canbe glued back together by Lemma 4.4 (b) to the long exact sequence (∗).

Example 4.6 (Exact sequence of a homomorphism). Let ϕ : M → N be a homomorphism of R-modules. By Example 4.3 (a) and (b) there are then short exact sequences

0−→ kerϕ −→Mϕ−→ imϕ −→ 0 and 0−→ imϕ −→ N −→ N/ imϕ −→ 0,

and hence we get a glued exact sequence

0−→ kerϕ −→Mϕ−→ N −→ N/ imϕ −→ 0

by Lemma 4.4 (b). So any homomorphism can be completed both to the left and to the right to anexact sequence with zero modules at the ends. Of course, the exactness of this sequence could alsobe checked directly, without using Lemma 4.4 (b).

06There is another much more subtle way to construct new exact sequences from old ones. This time,instead of gluing two sequences such that the ending module of the first sequence is the startingmodule of the second, we consider two short exact sequences such that one of them can be mappedto the other by a sequence of homomorphisms.

38 Andreas Gathmann

Lemma 4.7 (Snake Lemma). Let

M N P 0

0 M′ N′ P′

ϕ

α

ψ

β γ

ϕ ′ ψ ′

be a commutative diagram of R-modules with exact rows. Then there is a long exact sequence

kerα −→ kerβ −→ kerγ −→M′/ imα −→ N′/ imβ −→ P′/ imγ. (∗)

Moreover:

• if ϕ is injective, then so is the first map in the sequence (∗);• if ψ ′ is surjective, then so is the last map in the sequence (∗).

Remark 4.8. There is a nice graphical way to express the assertion of Lemma 4.7 as follows. Firstof all, if we complete the vertical homomorphisms α , β , and γ to exact sequences as in Example 4.6we obtain the solid arrows in the following diagram, in which all columns are exact.

0 0 0

0 kerα kerβ kerγ

0 M N P 0

0 M′ N′ P′ 0

M′/ imα N′/ imβ P′/ imγ 0

0 0 0

α β γ

The statement of the lemma is now the existence of an exact sequence indicated by the dashed arrowsabove. In particular, the probably unexpected homomorphism from kerγ to M′/ imα is the reasonfor the name “Snake Lemma”. The dotted arrows represent the additional statements of Lemma 4.7:one can read the diagram with or without both dotted arrows on the left, and with or without bothdotted arrows on the right.

Proof of Lemma 4.7. In the following proof, we will denote elements of M,N,P,M′,N′,P′ by thecorresponding lower case letters.

First of all let us construct the homomorphisms in the sequence

kerαϕ−→ kerβ

ψ−→ kerγδ−→M′/ imα

ϕ ′−→ N′/ imβψ ′−→ P′/ imγ. (∗)

(a) The first two maps. Let ϕ : kerα → kerβ , m 7→ ϕ(m) be the restriction of ϕ to kerα . Notethat its image lies indeed in kerβ : for m ∈ kerα we have β (ϕ(m)) = ϕ ′(α(m)) = 0 sincethe given diagram is commutative. In the same way, let ψ be the restriction of ψ to kerβ .

(b) The last two maps. We have ϕ ′(imα)⊂ imβ , since ϕ ′(α(m)) = β (ϕ(m)). Hence we get awell-defined map ϕ ′ : M′/ imα → N′/ imβ , m′ 7→ ϕ ′(m′). Similarly, set ψ ′(n′) := ψ ′(n′).

(c) The middle map δ . The existence of this map from a submodule of P to a quotient module ofM′ — it is usually called the connecting homomorphism — is the central part of this lemma.


Schematically, its idea is shown by the dashed arrow in the picture below: take an inverseimage under ψ , then the image under β , and then again an inverse image under ϕ ′.

M N P 0

0 M′ N′ P′

ϕ

α

ψ

β γ

ϕ ′ ψ ′

More precisely, let p∈ kerγ ≤P. As ψ is surjective there is an element n∈N with ψ(n) = p.Set n′ = β (n). Then ψ ′(n′) = ψ ′(β (n)) = γ(ψ(n)) = γ(p) = 0, so that n′ ∈ kerψ ′ = imϕ ′.Hence there is an element m′ ∈M′ with ϕ ′(m′) = n′. We want to set δ (p) := m′ ∈M′/ imα .

We have to check that this is well-defined, i. e. does not depend on the two choices of inverseimages. Note that the choice of m′ was clearly unique since ϕ ′ is injective. Now let ussuppose that we pick another inverse image n∈N of p in the first step, with image n′ = β (n)under β and inverse image m′ under ϕ ′. Then ψ(n− n) = 0, so n− n ∈ kerψ = imϕ , i. e.n−n = ϕ(m) for some m ∈M. It follows that

ϕ′(m′−m′) = n′−n′ = β (n−n) = β (ϕ(m)) = ϕ

′(α(m)),

and hence m′−m′ = α(m) since ϕ ′ is injective. Consequently, m′ and m′ define the sameclass in M′/ imα , and thus δ is in fact well-defined.

As all these maps are clearly homomorphisms, it only remains to prove that the sequence (∗) isexact. The technique to do this is the same as in the construction of the connecting homomorphismδ above: just following elements through the given commutative diagram, a so-called diagram chase.This is purely automatic and does not require any special ideas. The proof is therefore probably notvery interesting, but we will give it here nevertheless for the sake of completeness.

• Exactness at kerβ . Clearly, ψ(ϕ(m)) = ψ(ϕ(m)) = 0 for m ∈ kerα , and so im ϕ ⊂ ker ψ .Conversely, if n ∈ ker ψ ≤ N then n ∈ kerψ = imϕ , so n = ϕ(m) for some m ∈ M. Thiselement satisfies ϕ ′(α(m)) = β (ϕ(m)) = β (n) = 0 since n ∈ kerβ . By the injectivity of ϕ ′

this means that m lies in kerα , which is the source of ϕ . Hence n ∈ im ϕ .

• Exactness at kerγ . If p ∈ im ψ then p = ψ(n) for some n ∈ kerβ . In the construction (c) ofδ above, we can then choose this element n as inverse image for p, so we get n′ = β (n) = 0and thus δ (p) = 0, i. e. p ∈ kerδ . Conversely, if p ∈ kerδ then m′ = α(m) for some m ∈Min the construction of δ in (c). Continuing in the notation of (c), we then have β (n) =n′ = ϕ ′(m′) = ϕ ′(α(m)) = β (ϕ(m)). Hence, n−ϕ(m) ∈ kerβ with ψ(n−ϕ(m)) = ψ(n)−ψ(ϕ(m)) = ψ(n) = p, and thus p ∈ im ψ .

• Exactness at M′/ imα . If m′ ∈ imδ , then ϕ ′(m′) = n′ = 0 in the notation of (c) since n′ =β (n) ∈ imβ . Conversely, if m′ ∈ ker ϕ ′, then n′ := ϕ ′(m′) ∈ imβ , so n′ = β (n) for somen ∈ N. Then p := ψ(n) satisfies γ(p) = γ(ψ(n)) = ψ ′(β (n)) = ψ ′(n′) = ψ ′(ϕ ′(m′)) = 0and yields the given element m′ as image under δ , so m′ ∈ imδ .

• Exactness at N′/ imβ . If n′ ∈ im ϕ ′ then n′ = ϕ ′(m′) for some m′ ∈M′, and thus ψ ′(n′) =ψ ′(ϕ ′(m′)) = 0. Conversely, if n′ ∈ ker ψ ′ then ψ ′(n′) ∈ imγ , so ψ ′(n′) = γ(p) for somep ∈ P. As ψ is surjective, we can choose n ∈ N with ψ(n) = p. Then ψ ′(n′− β (n)) =γ(p)− γ(ψ(n)) = 0, and hence n′− β (n) ∈ kerψ ′ = imϕ ′. We therefore find an elementm′ ∈M′ with ϕ ′(m′) = n′−β (n), and thus ϕ ′(m′) = n′−β (n) = n′.

• Injectivity of ϕ . As ϕ is just a restriction of ϕ , it is clear that ϕ is injective if ϕ is.

• Surjectivity of ψ ′. If ψ ′ is surjective then for any p′ ∈ P′ there is an element n′ ∈ N′ withψ ′(n′) = p′, and thus ψ ′(n′) = p′. �

Exercise 4.9 (Hom( · ,N) is left exact).(a) Prove that a sequence

M1ϕ1−→M2

ϕ2−→M3 −→ 0

40 Andreas Gathmann

of R-modules is exact if and only if the sequence

0−→ Hom(M3,N)ϕ∗2−→ Hom(M2,N)

ϕ∗1−→ Hom(M1,N) (∗)

is exact for every R-module N, where ϕ∗i (ϕ) = ϕ ◦ϕi for i ∈ {1,2}.(b) Show that the statement of (a) is not true with an additional 0 at the left and right end,

respectively — i. e. that ϕ∗1 need not be surjective if ϕ1 is injective. We say that the operationHom( · ,N) is left exact (because the sequence (∗) is exact with 0 at the left), but not exact.

After having seen several ways to construct exact sequences, let us now study what sort of informa-tion we can get out of them. The general philosophy is that an exact sequence with zero modules atthe end is “almost enough” to determine any of its modules in terms of the others in the sequence.As a first statement in this direction, let us show that in order to compute the length of a moduleit is enough to have the module somewhere in an exact sequence in which the lengths of the othermodules are known — at least if all involved lengths are finite.

Corollary 4.10. Let

0−→M1ϕ1−→M2

ϕ2−→ ·· · ϕn−1−→Mn −→ 0

be an exact sequence of R-modules of finite length. Then ∑ni=1(−1)i l(Mi) = 0.

Proof. By Corollary 3.23 (b) applied to all morphisms ϕ1, . . . ,ϕn−1 and an index shift we get

n−1

∑i=1

(−1)i l(Mi) =n−1

∑i=1

(−1)i (l(kerϕi)+ l(imϕi))

=−l(kerϕ1)+(−1)n−1l(imϕn−1)+n−1

∑i=2

(−1)i (l(kerϕi)− l(imϕi−1)).

But l(kerϕ1) = 0 since ϕ1 is injective, l(imϕn−1) = l(Mn) since ϕn−1 is surjective, and l(kerϕi) =l(imϕi−1) for all i = 2, . . . ,n− 1 by the exactness of the sequence. Plugging this into the aboveformula, the result follows. �

Of course, knowing the length of a module does not mean that one knows the module up to isomor-phism (as we have seen e. g. in Example 3.21). So let us now address the question to what extent amodule in an exact sequence can be completely recovered from the other parts of the sequence. Forsimplicity, let us restrict to short exact sequences.

Example 4.11 (Recovering modules from an exact sequence). Consider an exact sequence

0−→Mϕ−→ N

ψ−→ P−→ 0

of R-modules, and let us ask whether we can determine one of the three modules if we know the restof the sequence.

Of course, if we know M and N together with the map ϕ then P is uniquely determined by this data:as we must have imψ = P and kerψ = imϕ , the homomorphism theorem of Proposition 3.10 (a)tells us that P∼= N/kerψ = N/ imϕ . In the same way, M is uniquely determined if we know N andP together with the map ψ , since by the injectivity of ϕ we have M ∼= imϕ = kerψ .

The most interesting question is thus if we can recover the middle term N if we know M and P (butnone of the maps). The following two examples show that this is in general not possible.

(a) In any case, a possible way to obtain a short exact sequence from given modules M at theleft and P at the right is

0−→Mϕ−→M⊕P

ψ−→ P−→ 0,

where ϕ is the inclusion of M in M⊕P, and ψ the projection of M⊕P onto P.


(b) There is an exact sequence of Z-modules

0−→ Z2·2−→ Z4 −→ Z2 −→ 0,

in which the morphism from Z4 to Z2 is just the quotient map. Note that in contrast to (a) itsmiddle term Z4 is not isomorphic to Z2⊕Z2 as a Z-module (i. e. as a group).

However, recovering the middle term in an exact sequence does work under a slightly strongerassumption: if we have two short exact sequences, and one of them maps to the other in the senseof the following lemma, then the two sequences must agree up to isomorphisms if they agree at anytwo of their modules.

Corollary 4.12 (5-Lemma). Let

0 M N P 0

0 M′ N′ P′ 0

ϕ

α

ψ

β γ

ϕ ′ ψ ′

be a commutative diagram of R-modules with exact rows. If two of the maps α , β , and γ areisomorphisms, then so is the third.

Proof. By the Snake Lemma 4.7 there is a long exact sequence

0−→ kerα −→ kerβ −→ kerγ −→M′/ imα −→ N′/ imβ −→ P′/ imγ −→ 0.

If now e. g. α and γ are bijective then kerα = kerγ = M′/ imα = P′/ imγ = 0, and the sequencebecomes

0−→ 0−→ kerβ −→ 0−→ 0−→ N′/ imβ −→ 0−→ 0.

By Example 4.2 (b) this means that kerβ = N′/ imβ = 0, and so β is bijective as well.

Of course, an analogous argument applies if α and β , or β and γ are assumed to be bijective. �

Remark 4.13. There are various versions of the 5-Lemma in the literature. In fact, looking at theproof above we see that our assumptions could be relaxed: in order to show that β is an isomorphismif α and γ are, we do not need the additional zero modules on the left and right of the long exactsequence of the Snake Lemma 4.7. So for this case we do not have to require ϕ to be injective orψ ′ to be surjective — this is only necessary if we want to conclude that α or γ is an isomorphism,respectively.

More generally, if one wants to show that the middle vertical map is an isomorphism, one can showthat it is enough to assume a commutative diagram

M1 M2 M3 M4 M5

M′1 M′2 M′3 M′4 M′5

β1 β2 β3 β4 β5

with exact rows in which the zero modules at the left and right ends have been replaced by arbitraryones, and where now β1, β2, β4, and β5 are required to be isomorphisms. This version of the lemmais actually the reason for the name “5-Lemma”.

Using Corollary 4.12, we can now give easy criteria to check whether a given short exact sequence0 −→M −→ N −→ P −→ 0 is of the simple form as in Example 4.11 (a), i. e. whether the middleentry N is just the direct sum of M and P:

Corollary 4.14 (Splitting Lemma). For a short exact sequence

0−→Mϕ−→ N

ψ−→ P−→ 0

the following three statements are equivalent:

42 Andreas Gathmann

(a) N ∼= M⊕P, with ϕ being the inclusion of M in the first summand, and ψ the projection ontothe second.

(b) ϕ has a left-sided inverse, i. e. there is a homomorphism α : N→M such that α ◦ϕ = idM .

(c) ψ has a right-sided inverse, i. e. there is a homomorphism β : P→ N such that ψ ◦β = idP.

In this case, the sequence is called split exact.

Proof.

(a) ⇒ (b) and (c): Under the assumption (a) we can take α to be the projection from M⊕P ontoM, and for β the inclusion of P in M⊕P.

(b) ⇒ (a): If α is a left-sided inverse of ϕ , the diagram

0 M N P 0

0 M M⊕P P 0

ϕ

id

ψ

(α,ψ) id

is commutative with exact rows. As the left and right vertical map are obviously isomor-phisms, so is the middle one by Corollary 4.12.

(c) ⇒ (a): If β is a right-sided inverse of ψ , the diagram

0 M M⊕P P 0

0 M N P 0

id ϕ+β idϕ ψ

is commutative with exact rows, where (ϕ +β )(m, p) := ϕ(m)+β (p). Again, as the leftand right vertical map are isomorphisms, so is the middle one by Corollary 4.12. �

Example 4.15. Every short exact sequence

0−→Uϕ−→V

ψ−→W −→ 0

of vector spaces over a field K is split exact: if (bi)i∈I is a basis of W we can pick inverse imagesci ∈ψ−1(bi) by the surjectivity of ψ . There is then a (unique) linear map β : W →V with β (bi) = ci[G2, Corollary 16.27]. Hence ψ ◦β = idW , i. e. the above sequence is split exact. So by Corollary4.14 we conclude that we always have V ∼=U⊕W in the above sequence.

Exercise 4.16. Let I and J be ideals in a ring R.

(a) Prove that there is an exact sequence of R-modules (what are the maps?)

0−→ I∩ J −→ I⊕ J −→ I + J −→ 0.

(b) Use the Snake Lemma 4.7 to deduce from this an exact sequence

0−→ R/(I∩ J)−→ R/I⊕R/J −→ R/(I + J)−→ 0.

(c) Show by example that the sequences of (a) and (b) are in general not split exact.

Exercise 4.17. Let N be a submodule of a finitely generated R-module M. In Exercise 3.5 (c) youhave seen that N need not be finitely generated in this case.

However, prove now that N is finitely generated if it is the kernel of a surjective R-module homo-morphism ϕ : M→ Rn for some n ∈ N.

(Hint: Show and use that the sequence 0−→ N −→Mϕ−→ Rn −→ 0 is split exact.)

5. Tensor Products 43

5. Tensor Products

In the last two chapters we have developed powerful methods to work with modules and linear mapsbetween them. However, in practice bilinear (or more generally multilinear) maps are often neededas well, so let us have a look at them now. Luckily, it turns out that to study bilinear maps we donot have to start from scratch, but rather can reduce their theory to the linear case. More precisely,for given R-modules M and N we will construct another module named M⊗N — the so-calledtensor product of M and N — such that bilinear maps from M×N to any other R-module P are innatural one-to-one correspondence with linear maps from M⊗N to P. So instead of bilinear mapsfrom M×N we can then always consider linear maps from the tensor product, and thus use all themachinery that we have developed so far for homomorphisms.

As the construction of this tensor product is a bit lengthy, let us first give an easy example that shouldshow the idea behind it.

Example 5.1 (Idea of tensor products). Let M and N be finitely generated free modules over a ringR, and choose bases B = (b1, . . . ,bm) and C = (c1, . . . ,cn) of M and N, respectively. Then everybilinear map α : M×N→ P to a third R-module P satisfies

α(λ1b1 + · · ·+λmbm,µ1c1 + · · ·+µncn) =m

∑i=1

n

∑j=1

λiµ j α(bi,c j) (∗)

for all λ1, . . . ,λm,µ1, . . . ,µn ∈ R. Hence α is uniquely determined by specifying the valuesα(bi,c j) ∈ P. Conversely, any choice of these values gives rise to a well-defined bilinear mapα : M×N→ P by the above formula.

Now let F be a free R-module of rank m ·n. We denote a basis of this space by bi⊗c j for i = 1, . . . ,mand j = 1, . . . ,n — so at this point this is just a name for a basis of this module, rather than anactual operation between elements of M and N. By the same argument as above, a linear mapϕ : F → P can then be specified uniquely by giving arbitrary images ϕ(bi⊗ c j) ∈ P of the basiselements [G2, Corollary 16.27]. Putting both results together, we see that there is a one-to-onecorrespondence between bilinear maps α : M×N → P and linear maps ϕ : F → P, given on thebases by α(bi,c j) = ϕ(bi⊗ c j): both maps can be specified by giving m ·n arbitrary elements of P.So in the above sense F is a tensor product of M and N.

If M and N are no longer free, but still finitely generated, we can at least pick generators (b1, . . . ,bm)and (c1, . . . ,cn) of M and N, respectively. Then (∗) shows that any bilinear map α : M×N → P isstill determined by the values α(bi,c j). But these values can no longer be chosen independently;they have to be compatible with the relations among the generators. For example, if we have therelation 2b1 + 3b2 = 0 in M, we must have 2α(b1,c j)+ 3α(b2,c j) = 0 for all j in order to get awell-defined bilinear map α . For the tensor product, this means the following: if G is the submoduleof F generated by all relations — so in our example we would take 2b1⊗ c j +3b2⊗ c j for all j —then bilinear maps α : M×N→ P now correspond exactly to those linear maps ϕ : F → P that arezero on G. As these are the same as linear maps from F/G to P, we can now take F/G to be ourtensor product of M and N.

In fact, this idea of the construction of the tensor product should be clearly visible in the proofof Proposition 5.5 below. The main difference will be that, in order to avoid unnatural choices ofgenerators, we will just take all elements of M and N as a generating set. This will lead to a hugemodule F , but also to a huge submodule G of relations among these generators, and so the quotientF/G will again be what we want.

07But let us now see how to obtain the tensor product M⊗N rigorously. There are two options for this:we can either construct it directly and then prove its properties, or define it to be a module having the

44 Andreas Gathmann

desired property — namely that linear maps from M⊗N are the same as bilinear maps from M×N— and show that such an object exists and is uniquely determined by this property. As this propertyis actually much more important than the technical construction of M⊗N, we will take the latterapproach.

Definition 5.2 (Tensor products). Let M, N, and P be R-modules.

(a) A map α : M×N → P is called R-bilinear if α( · ,n) : M → P and α(m, ·) : N → P areR-linear for all m ∈M and n ∈ N.

(b) A tensor product of M and N over R is an R-module T togetherwith a bilinear map τ : M×N→ T such that the following universalproperty holds: for every bilinear map α : M×N → P to a thirdmodule P there is a unique linear map ϕ : T → P such that α = ϕ ◦τ ,i. e. such that the diagram on the right commutes. The elements of atensor product are called tensors.

M×N P

T

α

τϕ

Remark 5.3. In the above notation, Definition 5.2 (b) just means that there is a one-to-one corre-spondence

{bilinear maps M×N→ P} 1:1←→ {homomorphisms T → P}

α 7−→ ϕ

ϕ ◦ τ ←−7 ϕ

as explained in the motivation above.

Proposition 5.4 (Uniqueness of tensor products). A tensor product is uniqueup to unique isomorphism in the following sense: if T1 and T2 together withbilinear maps τ1 : M×N→ T1 and τ2 : M×N→ T2 are two tensor productsfor M and N over R, there is a unique R-module isomorphism ϕ : T1 → T2such that τ2 = ϕ ◦ τ1.

M×N T2

T1

τ2

τ1ϕ

Proof. Consider the universal property of Definition 5.2 (b) for the first tensor product: as τ2 : M×N → T2 is bilinear, there is a unique morphism ϕ : T1 → T2 with τ2 = ϕ ◦ τ1. In the same way,reversing the roles of the tensor products we get a unique morphism ψ : T2→ T1 with τ1 = ψ ◦ τ2.

Now apply the universal property for the first tensor product again, this timefor the bilinear map τ1 : M×N → T1 as shown on the right. Note that wehave ψ ◦ϕ ◦ τ1 = ψ ◦ τ2 = τ1 as well as idT1 ◦τ1 = τ1, so that both ψ ◦ϕ

and idT1 make the diagram commute. Hence, by the uniqueness part of theuniversal property we conclude that ψ ◦ϕ = idT1 . In the same way we seethat ϕ ◦ψ = idT2 , and thus ϕ is an isomorphism.

M×N T1

T1

τ1

τ1id

ψ ◦ϕ

�

Proposition 5.5 (Existence of tensor products). Any two R-modules have a tensor product.

Proof. Let M and N be R-modules. We denote by F the R-module of all finite formal linear combi-nations of elements of M×N, i. e. formal sums of the form

a1 (m1,n1)+ · · ·+ak (mk,nk)

for k ∈ N, a1, . . . ,ak ∈ R, and distinct (mi,ni) ∈ M×N for i = 1, . . . ,k. More precisely, F can bemodeled as the set of maps from M×N to R that have non-zero values at most at finitely manyelements, where the values at these elements (m1,n1), . . . ,(mk,nk) ∈ M×N are a1, . . . ,ak in theabove notation. In this picture, the R-module structure of F is then given by pointwise addition andscalar multiplication. It is more intuitive however to think of the elements of F as linear combinationsof elements of M×N as above (rather than as functions from M×N to R), and so we will use thisnotation in the rest of the proof.

Note that the various elements (m,n) ∈ M×N are by definition all independent in F — e. g. forgiven a ∈ R, m ∈ M, and n ∈ N the linear combinations a(m,n), 1(am,n), and 1(m,an) are in


general all different elements of F . In order to construct the tensor product we now want to enforcejust enough relations so that the formal linear combinations become bilinear: let G be the submoduleof F generated by all expressions

(m1 +m2,n)− (m1,n)− (m2,n), (am,n)−a(m,n),

(m,n1 +n2)− (m,n1)− (m,n2), and (m,an)−a(m,n)

for all a ∈ R, m,m1,m2 ∈M, and n,n1,n2 ∈ N, and set T := F/G. Then the map

τ : M×N→ T, (m,n) 7→ (m,n)

is R-bilinear by the very definition of these relations.

We claim that T together with τ is a tensor product for M and N over R. So to check the universalproperty let α : M×N→ P be an R-bilinear map. Then we can define a homomorphism ϕ : T → Pby setting ϕ

((m,n)

):= α(m,n) and extending this by linearity, i. e.

ϕ(a1 (m1,n1)+ · · ·+ak (mk,nk)

)= a1 α(m1,n1)+ · · ·+ak α(mk,nk).

Note that ϕ is well-defined since α is bilinear, and we certainly have α = ϕ ◦ τ . Moreover, it is alsoobvious that setting ϕ

((m,n)

)= α(m,n) is the only possible choice such that α = ϕ ◦ τ . Hence the

universal property is satisfied, and T together with τ is indeed a tensor product. �

Notation 5.6 (Tensor products). Let M and N be R-modules. By Propositions 5.4 and 5.5 there isa unique tensor product of M and N over R up to isomorphism, i. e. an R-module T together witha bilinear map τ : M×N → T satisfying the universal property of Definition 5.2 (b). We write Tas M⊗R N (or simply M⊗N if the base ring is understood), and τ(m,n) as m⊗ n. The elementm⊗n ∈M⊗N is often called the tensor product of m and n.

Remark 5.7.(a) Tensors in M ⊗N that are of the form m⊗ n for m ∈ M and n ∈ N are called pure or

monomial. As we can see from the proof of Proposition 5.5, not every tensor in M⊗Nis pure — instead, the pure tensors generate M⊗N as an R-module, i. e. a general element ofM⊗N can be written as a finite linear combination ∑

ki=1 ai (mi⊗ni) for k ∈N, a1, . . . ,ak ∈ R,

m1, . . . ,mk ∈M, and n1, . . . ,nk ∈ N.

Note that these generators are not independent, so that there are in general many differentways to write a tensor as a linear combination of pure tensors. This makes it often a non-trivial task to decide whether two such linear combinations are the same tensor or not.

(b) The tensor product of two elements of M and N is bilinear by Definition 5.2 (b), i. e. we have

(m1 +m2)⊗n = m1⊗n+m2⊗n and a(m⊗n) = (am)⊗n

in M⊗N for all a ∈ R, m,m1,m2 ∈ M, and n ∈ N, and similarly for the second factor. Infact, the tensor product has been defined in such a way that the relations among tensors areexactly those bilinear ones.

(c) Of course, using multilinear instead of bilinear maps, we can also define tensor productsM1⊗·· ·⊗Mk of more than two modules in the same way as above. We will see in Exercise5.9 however that the result is nothing but a repeated application of the tensor product forbilinear maps.

Before we give some examples of tensor product spaces, let us first prove a few simple propertiesthat will also make the study of the examples easier.

Lemma 5.8. For any R-modules M, N, and P there are natural isomorphisms

(a) M⊗N ∼= N⊗M;

(b) M⊗R∼= M;

(c) (M⊕N)⊗P∼= (M⊗P)⊕ (N⊗P).

46 Andreas Gathmann

Proof. The strategy for all three proofs is the same: using the universal property of Definition 5.2(b) we construct maps between the tensor products from bilinear maps, and then show that they areinverse to each other.

(a) The map M×N→ N⊗M, (m,n) 7→ n⊗m is bilinear by Remark 5.7 (b), and thus induces a(unique) linear map

ϕ : M⊗N→ N⊗M with ϕ(m⊗n) = n⊗m for all m ∈M and n ∈ N

by the universal property of M⊗N. In the same way we get a morphism ψ : N⊗M→M⊗Nwith ψ(n⊗m) = m⊗ n. Then (ψ ◦ϕ)(m⊗ n) = m⊗ n for all m ∈ M and n ∈ N, so ψ ◦ϕ

is the identity on pure tensors. But the pure tensors generate M⊗N, and so we must haveψ ◦ϕ = idM⊗N . In the same way we see that ϕ ◦ψ = idN⊗M . Hence ϕ is an isomorphism.

(b) From the bilinear map M×R→M, (m,a) 7→ am we obtain a linear map

ϕ : M⊗R→M with ϕ(m⊗a) = am for all m ∈M and a ∈ R

by the universal property. Furthermore, there is a linear map ψ : M→M⊗R, m 7→ m⊗ 1.As

(ψ ◦ϕ)(m⊗a) = am⊗1 = m⊗a and (ϕ ◦ψ)(m) = m

for all m ∈M and a ∈ R, we conclude as in (a) that ϕ is an isomorphism.

(c) The bilinear map (M⊕N)×P→ (M⊗P)⊕ (N⊗P), ((m,n), p) 7→ (m⊗ p,n⊗ p) induces alinear map

ϕ : (M⊕N)⊗P→ (M⊗P)⊕ (N⊗P) with ϕ((m,n)⊗ p) = (m⊗ p,n⊗ p)

as above. Similarly, we get morphisms M⊗P→ (M⊕N)⊗P with m⊗ p 7→ (m,0)⊗ p andN⊗P→ (M⊕N)⊗P with n⊗ p 7→ (0,n)⊗ p, and thus by addition a linear map

ψ : (M⊗P)⊕ (N⊗P)→ (M⊕N)⊗P with ψ(m⊗ p,n⊗q) = (m,0)⊗ p+(0,n)⊗q.

Is is verified immediately that ϕ and ψ are inverse to each other on pure tensors, and thusalso on the whole tensor product space. �

Exercise 5.9 (Associativity of tensor products). Let M, N, and P be three R-modules. Prove thatthere are natural isomorphisms

M⊗N⊗P∼= (M⊗N)⊗P∼= M⊗ (N⊗P)

where M⊗N⊗P is the tensor product constructed from trilinear maps as in Remark 5.7 (c).

Example 5.10.(a) Let M and N be free R-modules of ranks m and n, respectively. Then M ∼= Rm and N ∼= Rn,

and so by Lemma 5.8 we have

M⊗N ∼= Rm⊗ (R⊕·· ·⊕R︸︷︷︸n times

)∼= (Rm⊗R)⊕·· ·⊕ (Rm⊗R)∼= Rm⊕·· ·⊕Rm ∼= Rmn,

as expected from Example 5.1. So after a choice of bases the elements of M⊗N can bedescribed by m×n-matrices over R.

(b) Let I and J be coprime ideals in a ring R. Then there are elements a ∈ I and b ∈ J witha+ b = 1, and so we obtain in the tensor product R/I⊗R/J for all monomial tensors r⊗ swith r,s ∈ R

r⊗ s = (a+b)(r⊗ s) = ar⊗ s+ r⊗bs = 0

since ar = 0 ∈ R/I and bs = 0 ∈ R/J. But these monomial tensors generate the tensorproduct, and so we conclude that R/I⊗R/J = 0. As a concrete example, we get Zp⊗ZZq = 0for any two distinct primes p and q.

This shows that (in contrast to (a)) a tensor product space need not be “bigger” than itsfactors, and might be 0 even if none of its factors are.


(c) In the tensor product Z⊗ZZ2 we have

2⊗1 = 1⊗2 = 0

by bilinearity. However, if we now consider 2⊗ 1 as an element of 2Z⊗Z Z2, the abovecomputation is invalid since 1 /∈ 2Z. So is it still true that 2⊗1 = 0 in 2Z⊗ZZ2?

We can answer this question with Lemma 5.8 (b): we know that 2Z is isomorphic to Zas a Z-module by sending 2 to 1, and so 2Z⊗Z Z2 is isomorphic to Z⊗Z Z2 ∼= Z2 by themap ϕ : 2Z⊗Z Z2→ Z2, 2a⊗ b 7→ ab. But now ϕ(2⊗ 1) = 1 6= 0, and so indeed we have2⊗1 6= 0 in 2Z⊗Z2.

The conclusion is that, when writing down tensor products m⊗n, we have to be very carefulto specify which tensor product space we consider: if n ∈ N and m lies in a submodule M′

of a module M, it might happen that m⊗ n is non-zero in M′⊗N, but zero in M⊗N. Inother words, for a submodule M′ of M it is not true in general that M′⊗N is a submodule ofM⊗N in a natural way! We will discuss this issue in more detail in Proposition 5.22 (b) andRemark 5.23.

Exercise 5.11. Compute the tensor products Q/Z⊗ZQ/Z, Q⊗ZQ, C⊗RC, and Q[x]⊗QC.

Exercise 5.12. Assume that we have n rectangles R1, . . . ,Rn in theplane, of size ai × bi for i = 1, . . . ,n, that fit together to form arectangle R of size a×b as in the picture on the right. Prove:

(a) a⊗b = ∑ni=1 ai⊗bi in R⊗QR.

(b) If each of the rectangles R1, . . . ,Rn has at least one sidewith a rational length, then R must also have at least oneside with a rational length. a

bR1

R2

R3R4

R5

Exercise 5.13 (Dual vector spaces). Let V and W be vector spaces over a field K. We call V ∗ :=HomK(V,K) the dual vector space to V . Moreover, denote by BLF(V ) the vector space of bilinearforms V ×V → K.

(a) Show that there are (natural, i. e. basis-independent) linear maps

Φ : V ∗⊗W → Hom(V,W ) such that Φ(ϕ⊗w)(v) = ϕ(v) ·w,Ψ : V ∗⊗V ∗→ BLF(V ) such that Ψ(ϕ⊗ψ)(v,v′) = ϕ(v) ·ψ(v′),

T : V ∗⊗V → K such that T (ϕ⊗ v) = ϕ(v).

(b) Prove that Φ and Ψ are injective.

(c) Prove that Φ and Ψ are in fact isomorphisms if V and W are finite-dimensional, but not ingeneral for arbitrary vector spaces.

(d) Assume that dimK V = n < ∞, so that V ∗⊗V is naturally isomorphic to HomK(V,V ) by (c),which in turn is isomorphic to Mat(n× n,K) in the standard way after choosing a basis ofV . Using these isomorphisms, T becomes a linear map Mat(n×n,K)→ K that is invariantunder a change of basis. Which one?

08To define homomorphisms between tensor products the following construction will be useful.

Construction 5.14 (Tensor product of homomorphisms). Let ϕ : M → N and ϕ ′ : M′ → N′ betwo homomorphisms of R-modules. Then the map M×M′→ N⊗N′, (m,m′) 7→ ϕ(m)⊗ϕ ′(m′) isbilinear, and thus by the universal property of the tensor product gives rise to a homomorphism

ϕ⊗ϕ′ : M⊗M′→ N⊗N′ such that (ϕ⊗ϕ

′)(m⊗m′) = ϕ(m)⊗ϕ′(m′)

for all m ∈M and m′ ∈M′. We call ϕ⊗ϕ ′ the tensor product of ϕ and ϕ ′.

48 Andreas Gathmann

Remark 5.15. Note that for ϕ ∈ HomR(M,N) and ϕ ′ ∈ HomR(M′,N′) we have already defined atensor product ϕ ⊗ϕ ′ as an element of HomR(M,N)⊗R HomR(M′,N′) in Notation 5.6 — whereasConstruction 5.14 gives an element of HomR(M⊗R M′,N⊗R N′). In fact, it is easy to see by theuniversal property of the tensor product that there is a natural homomorphism

HomR(M,N)⊗R HomR(M′,N′)→ HomR(M⊗R M′,N⊗R N′)

that sends ϕ ⊗ ϕ ′ in the sense of Notation 5.6 to the morphism of Construction 5.14. It shouldtherefore not lead to confusion if we denote both constructions by ϕ⊗ϕ ′.

One application of tensor products is to extend the ring of scalars for a given module. For vectorspaces, this is a process that you know very well: suppose that we have e. g. a real vector space Vwith dimRV = n < ∞ and want to study the eigenvalues and eigenvectors of a linear map ϕ : V →V .We then usually set up the matrix A ∈Mat(n×n,R) corresponding to ϕ in some chosen basis, andcompute its characteristic polynomial. Often it happens that this polynomial does not split into linearfactors over R, and that we therefore want to pass from the real to the complex numbers.

But while it is perfectly possible to consider A as a complex matrix in Mat(n×n,C) instead and talkabout complex eigenvalues and eigenvectors of A, it is not clear what this means in the language ofthe linear map ϕ: in the condition ϕ(x) = λx for an eigenvector of ϕ it certainly does not make senseto take x to be a “complex linear combination” of the basis vectors, since such an element does notexist in V , and so ϕ is not defined on it. We rather have to extend V first to a complex vector space,and ϕ to a C-linear map on this extension. It turns out that the tensor product of V with C over R isexactly the right construction to achieve this in a basis-independent language.

Construction 5.16 (Extension of scalars). Let M be an R-module, and R′ an R-algebra (so that R′

is a ring as well as an R-module). Moreover, for any a ∈ R′ we denote by µa : R′→ R′, s 7→ as themultiplication map, which is obviously a homomorphism of R-modules. If we then set MR′ := M⊗RR′, we obtain a scalar multiplication with R′ on MR′

R′×MR′ → MR′

(a,m⊗ s) 7→ a · (m⊗ s) := (1⊗µa)(m⊗ s) = m⊗ (as)

which turns MR′ into an R′-module. We say that MR′ is obtained from M by an extension of scalarsfrom R to R′. Note that any R-module homomorphism ϕ : M→ N then gives rise to an “extended”R′-module homomorphism ϕR′ := ϕ⊗ id : MR′ → NR′ .

Example 5.17 (Complexification of a real vector space). Let V be a real vector space. Extendingscalars from R to C, we call VC = V ⊗RC the complexification of V . By Construction 5.16, this isnow a complex vector space.

Let us assume for simplicity that V is finitely generated, with basis (b1, . . . ,bn). Then V ∼= Rn by anisomorphism that maps bi to the i-th standard basis vector ei for i = 1, . . . ,n, and consequently

VC = V ⊗RC ∼= Rn⊗RC ∼= (R⊗RC)n ∼= Cn

as R-modules by Lemma 5.8. But by definition of the scalar multiplication with C from Construction5.16 this is also a C-module homomorphism, and thus an isomorphism of complex vector spaces.Moreover, bi⊗1 maps to ei under this chain of isomorphisms for all i, and so as expected the vectorsb1⊗1, . . . ,bn⊗1 form a basis of the complexified vector space VC.

Finally, let us consider a linear map ϕ : V → V described by the matrix A ∈ Mat(n× n,R) withrespect to the basis (b1, . . . ,bn), i. e. we have ϕ(bi) = ∑

nj=1 a j,ib j for all i. Then ϕC : VC→VC from

Construction 5.16 is an endomorphism of the complexified vector space, and since

ϕC(bi⊗1) = ϕ(bi)⊗1 =n

∑j=1

a j,i (b j⊗1)

we see that the matrix of ϕC with respect to the basis (b1⊗ 1, . . . ,bn⊗ 1) is precisely A again, justnow considered as a complex matrix.


Of course, the same constructions can not only be used to pass from the real to the complex numbers,but also for any field extension.

Exercise 5.18. Let M and N be R-modules, and R′ an R-algebra. Show that the extension of scalarscommutes with tensor products in the sense that there is an isomorphism of R′-modules

(M⊗R N)R′∼= MR′ ⊗R′ NR′ .

In Construction 5.16 we have considered the case when one of the two factors in a tensor productover a ring R is not only an R-module, but also an R-algebra. If both factors are R-algebras, we cansay even more: in this case, the resulting tensor product will also be an R-algebra in a natural way:

Construction 5.19 (Tensor product of algebras). Let R be a ring, and let R1 and R2 be R-algebras.Then the map R1 × R2 × R1 × R2 → R1 ⊗ R2, (s, t,s′, t ′) 7→ (ss′)⊗ (tt ′) is multilinear, and so bythe universal property of the tensor product (and its associativity as in Exercise 5.9) it induces ahomomorphism

(R1⊗R2)⊗ (R1⊗R2)→ R1⊗R2 with (s⊗ t)⊗ (s′⊗ t ′) 7→ (ss′)⊗ (tt ′)

for all s,s′ ∈ R1 and t, t ′ ∈ R2. Again by the universal property of the tensor product this nowcorresponds to a bilinear map

(R1⊗R2)× (R1⊗R2)→ R1⊗R2 with (s⊗ t,s′⊗ t ′) 7→ (s⊗ t) · (s′⊗ t ′) := (ss′)⊗ (tt ′).

It is obvious that this multiplication makes R1⊗R2 into a ring, and thus into an R-algebra. So thetensor product of two R-algebras has again a natural structure of an R-algebra.

Example 5.20 (Multivariate polynomial rings as tensor products). Let R be a ring. We claim thatR[x,y]∼= R[x]⊗R R[y] as R-algebras, i. e. that polynomial rings in several variables can be thought ofas tensor products of polynomial rings in one variable.

In fact, there are R-module homomorphisms

ϕ : R[x]⊗R R[y]→ R[x,y], f ⊗g 7→ f g

(by the universal property of the tensor product) and

ψ : R[x,y] 7→ R[x]⊗R[y], ∑i, j

ai, jxiy j 7→∑i, j

ai, jxi⊗ y j.

As

(ψ ◦ϕ)(xi⊗ y j) = ψ(xiy j) = xi⊗ y j

and (ϕ ◦ψ)(xiy j) = ϕ(xi⊗ y j) = xiy j

for all i, j ∈ N and these elements xi⊗ y j and xiy j generate R[x]⊗R R[y] and R[x,y] as an R-module,respectively, we see that ϕ and ψ are inverse to each other. Moreover, ϕ is also a ring homomorphismwith the multiplication in R[x]⊗R R[y] of Construction 5.19, since

ϕ(( f ⊗g) · ( f ′⊗g′)) = ϕ(( f f ′)⊗ (gg′)) = f f ′gg′ = ϕ( f ⊗g) ·ϕ( f ′⊗g′).

Hence R[x,y]∼= R[x]⊗R R[y] as R-algebras.

Exercise 5.21.(a) Let I and J be ideals in a ring R. Prove that R/I⊗R R/J ∼= R/(I + J) as R-algebras.

(b) Let X ⊂AnK and Y ⊂Am

K be varieties over a field K, so that X×Y ⊂ Kn+m. Show that X×Yis again a variety, and A(X×Y )∼= A(X)⊗K A(Y ) as K-algebras.

Finally, to conclude this chapter we want to study how tensor products behave in exact sequences.The easiest way to see this is to trace it back to Exercise 4.9, in which we applied HomR( ·N) to anexact sequence.

Proposition 5.22 (Tensor products are right exact).(a) For any R-modules M, N, and P we have Hom(M,Hom(N,P))∼= Hom(M⊗N,P).

50 Andreas Gathmann

(b) Let

M1ϕ1−→M2

ϕ2−→M3 −→ 0

be an exact sequence of R-modules. Then for any R-module N the sequence

M1⊗Nϕ1⊗id−→ M2⊗N

ϕ2⊗id−→ M3⊗N −→ 0

is exact as well.

Proof.

(a) A natural isomorphism can be constructed by identifying α ∈ Hom(M,Hom(N,P)) withβ ∈ Hom(M⊗N,P), where

α(m)(n) = β (m⊗n) ∈ P

for all m ∈ M and n ∈ N. Note that this equation can be used to define α in terms of β toget a map Hom(M⊗N,P)→ Hom(M,Hom(N,P)), and also (by the universal property ofthe tensor product) to define β in terms of α in order to get a map in the opposite direction.Obviously, these two maps are then R-linear and inverse to each other.

(b) Starting from the given sequence, Exercise 4.9 (a) gives us an exact sequence

0−→ Hom(M3,Hom(N,P))−→ Hom(M2,Hom(N,P))−→ Hom(M1,Hom(N,P))

for all R-modules N and P, where the two non-trivial maps send αi ∈ Hom(Mi,Hom(N,P))to αi−1 ∈ Hom(Mi−1,Hom(N,P)) with αi−1(mi−1)(p) = αi(ϕi−1(mi−1))(p) for i ∈ {2,3}and all m1 ∈M1, m2 ∈M2, p ∈ P. Using the isomorphism of (a), this is the same as an exactsequence

0−→ Hom(M3⊗N,P)−→ Hom(M2⊗N,P)−→ Hom(M1⊗N,P),

where the maps are now βi 7→ βi−1 with

βi−1(mi−1⊗ p) = βi(ϕi−1(mi−1)⊗ p) = βi((ϕi−1⊗ id)(mi−1⊗ p))

for i ∈ {2,3}. But using Exercise 4.9 (a) again, this means that the sequence

M1⊗Nϕ1⊗id−→ M2⊗N

ϕ2⊗id−→ M3⊗N −→ 0

is also exact. �

Remark 5.23. Similarly to the case of Hom( · ,N) in Exercise 4.9, the statement of Proposition 5.22(b) is in general not true with an additional zero module at the left, i. e. if ϕ1 is injective it does notnecessarily follow that ϕ1⊗ id is injective. In fact, we know this already from Example 5.10 (c),where we have seen that for a submodule M1 of M2 we cannot conclude that M1⊗N is a submoduleof M2⊗N in a natural way.

In analogy to Exercise 4.9 we say that taking tensor products is right exact, but not exact. However,we have the following result:

Exercise 5.24. Let 0−→M1 −→M2 −→M3 −→ 0 be a short exact sequence of R-modules. Showthat the sequence 0−→M1⊗N −→M2⊗N −→M3⊗N −→ 0 is also exact if one of the followingassumptions hold:

(a) the sequence 0−→M1 −→M2 −→M3 −→ 0 is split exact;

(b) N is a finitely generated, free R-module.

As an example of how Proposition 5.22 can be used, let us prove the following statement.

Corollary 5.25. Let I be an ideal in a ring R, and let M be an R-module. Then M/IM ∼= M⊗R R/I.


Proof. By Example 4.3 (b) we know that the sequence

0−→ I −→ R−→ R/I −→ 0

is exact. So by Proposition 5.22 (b) it follows that

M⊗R Iϕ−→M⊗R R

ψ−→M⊗R R/I −→ 0

is also exact, where ψ is the projection in the second factor, M⊗R R ∼= M by Lemma 5.8 (b), andϕ(m⊗a) = am using this isomorphism. In particular, the image of ϕ is just IM by Definition 3.12(a), and so we conclude by the exactness of the sequence and the homomorphism theorem

M⊗R R/I = imψ ∼= M/kerψ = M/ imϕ = M/IM. �

Example 5.26 (Derivatives in terms of tensor products). Let V and W be normed real vector spaces[G2, Definition 23.1], and let f : U →W be a function on an open subset U ⊂ V . Then for anypoint a ∈U the derivative f ′(a) of f in a (if it exists) is an element of HomR(V,W ): it is just thehomomorphism such that x 7→ f (a)+ f ′(a)(x−a) is the affine-linear approximation of f at a [G2,Definition 25.3 and Remark 25.6].

If we now want to define the second derivative f ′′ of f , the most natural way to do this isto take the derivative of the map f ′ : U → HomR(V,W ), a 7→ f ′(a), with a suitable norm onHomR(V,W ). By the same reasoning as above, this will now lead to an element f ′′(a) ofHomR(V,HomR(V,W )) [G2, Remark 26.5 (a)]. Similarly, the third derivative f ′′′(a) is an elementof HomR(V,HomR(V,HomR(V,W ))), and so on.

With Proposition 5.22 (a) we can now rephrase this in a simpler way in terms of tensor products: thek-th derivative f (k)(a) of f in a point a ∈U is an element of HomR(V ⊗R · · ·⊗RV,W ) for k ∈ N>0,where we take the k-fold tensor product of V with itself. So the higher derivatives of f can again bethought of as linear maps, just with a tensor product source space. Of course, if V and W are finite-dimensional with n = dimRV and m = dimRW , then HomR(V ⊗R · · ·⊗RV,W ) is of dimension nkm,and the coordinates of f (k)(a) with respect to bases of V and W are simply the nk partial derivativesof order k of the m coordinate functions of f .

Exercise 5.27. Show that l(M⊗N) ≤ l(M) · l(N) for any two R-modules M and N (where an ex-pression 0 ·∞ on the right hand side is to be interpreted as 0). Does equality hold in general?

(Hint: It is useful to consider suitable exact sequences.)09

52 Andreas Gathmann

6. Localization

Localization is a very powerful technique in commutative algebra that often allows to reduce ques-tions on rings and modules to a union of smaller “local” problems. It can easily be motivated bothfrom an algebraic and a geometric point of view, so let us start by explaining the idea behind it inthese two settings.

Remark 6.1 (Motivation for localization).(a) Algebraic motivation: Let R be a ring which is not a field, i. e. in which not all non-zero

elements are units. The algebraic idea of localization is then to make more (or even all)non-zero elements invertible by introducing fractions, in the same way as one passes fromthe integers Z to the rational numbers Q.

Let us have a more precise look at this particular example: in order to construct the rationalnumbers from the integers we start with R = Z, and let S = Z\{0} be the subset of theelements of R that we would like to become invertible. On the set R×S we then consider theequivalence relation

(a,s)∼ (a′,s′) ⇔ as′−a′s = 0

and denote the equivalence class of a pair (a,s) by as . The set of these “fractions” is then

obviously Q, and we can define addition and multiplication on it in the expected way byas +

a′s′ := as′+a′s

ss′ and as ·

a′s′ := aa′

ss′ .

(b) Geometric motivation: Now let R = A(X) be the ring of polynomial functions on a varietyX . In the same way as in (a) we can ask if it makes sense to consider fractions of suchpolynomials, i. e. rational functions

X → K, x 7→ f (x)g(x)

for f ,g ∈ R. Of course, for global functions on all of X this does not work, since g willin general have zeroes somewhere, so that the value of the rational function is ill-defined atthese points. But if we consider functions on subsets of X we can allow such fractions. Themost important example from the point of view of localization is the following: for a fixedpoint a ∈ X let S = { f ∈ A(X) : f (a) 6= 0} be the set of all polynomial functions that do notvanish at a. Then the fractions f

g for f ∈ R and g ∈ S can be thought of as rational functionsthat are well-defined at a.

In fact, the best way to interpret the fractions fg with f ∈ R and g ∈ S is as “functions on

an arbitrarily small neighborhood of a”, assuming that we have a suitable topology on X :although the only point at which these functions are guaranteed to be well-defined is a, forevery such function there is a neighborhood on which it is defined as well, and to a certainextent the polynomials f and g can be recovered from the values of the function on such aneighborhood. In this sense we will refer to functions of the form f

g with f ∈ R and g ∈ Sas “local functions” at a. In fact, this geometric interpretation is the origin of the name“localization” for the process of constructing fractions described in this chapter.

Let us now introduce such fractions in a rigorous way. As above, R will always denote the originalring, and S ⊂ R the set of elements that we would like to become invertible. Note that this subset Sof denominators for our fractions has to be closed under multiplication, for otherwise the formulasas +

a′s′ := as′+a′s

ss′ and as ·

a′s′ := aa′

ss′ for addition and multiplication of fractions would not make sense.Moreover, the equivalence relation on R×S to obtain fractions will be slightly different from the onein Remark 6.1 (a) — we will explain the reason for this later in Remark 6.3.

6. Localization 53

Lemma and Definition 6.2 (Localization of rings). Let R be a ring.

(a) A subset S⊂ R is called multiplicatively closed if 1 ∈ S, and ab ∈ S for all a,b ∈ S.

(b) Let S⊂ R be a multiplicatively closed set. Then

(a,s)∼ (a′,s′) :⇔ there is an element u ∈ S such that u(as′−a′s) = 0

is an equivalence relation on R×S. We denote the equivalence class of a pair (a,s) ∈ R×Sby a

s . The set of all equivalence classes

S−1R :={a

s: a ∈ R,s ∈ S

}is called the localization of R at the multiplicatively closed set S. It is a ring together withthe addition and multiplication

as+

a′

s′:=

as′+a′sss′

andas· a′

s′:=

aa′

ss′.

Proof. The relation ∼ is clearly reflexive and symmetric. It is also transitive: if (a,s)∼ (a′,s′) and(a′,s′)∼ (a′′,s′′) there are u,v ∈ S with u(as′−a′s) = v(a′s′′−a′′s′) = 0, and thus

s′uv(as′′−a′′s) = vs′′ ·u(as′−a′s)+us · v(a′s′′−a′′s′) = 0. (∗)

But this means that (a,s)∼ (a′′,s′′), since S is multiplicatively closed and therefore s′uv ∈ S.

In a similar way we can check that the addition and multiplication in S−1R are well-defined: if(a,s)∼ (a′′,s′′), i. e. u(as′′−a′′s) = 0 for some u ∈ S, then

u((as′+a′s)(s′′s′)− (a′′s′+a′s′′)(ss′)

)= (s′)2 ·u(as′′−a′′s) = 0

and u((aa′)(s′′s′)− (a′′a′)(ss′)

)= a′s′ ·u(as′′−a′′s) = 0,

and so as′+a′sss′ = a′′s′+a′s′′

s′′s′ and aa′ss′ =

a′′a′s′′s′ .

It is now verified immediately that these two operations satisfy the ring axioms — in fact, the re-quired computations to show this are exactly the same as for ordinary fractions in Q. �

Remark 6.3 (Explanation for the equivalence relation in Definition 6.2 (b)). Compared to Remark6.1 (a) it might be surprising that the equivalence relation (a,s) ∼ (a′,s′) for S−1R is not just givenby as′− a′s = 0, but by u(as′− a′s) = 0 for some u ∈ S. Again, there is both an algebraic and ageometric reason for this:

(a) Algebraic reason: Without the additional element u ∈ S the relation ∼ would not be transi-tive. In fact, if we set u = v = 1 in the proof of transitivity in Lemma 6.2, we could onlyshow that s′(as′′− a′′s) = 0 in (∗). Of course, this does not imply as′′− a′′s = 0 if S con-tains zero-divisors. (On the other hand, if S does not contain any zero-divisors, the conditionu(as′−a′s) = 0 for some u ∈ S can obviously be simplified to as′−a′s = 0.)

(b) Geometric reason, continuing Remark 6.1 (b): Let X = V (xy)in A2

R be the union of the two coordinate axes as in the pictureon the right, so that A(X) = R[x,y]/(xy). Then the functions yand 0 on X agree in a neighborhood of the point a = (1,0), andso y

1 and 01 should be the same local function at a, i. e. the same

element in S−1R with S= { f ∈A(X) : f (a) 6= 0}. But without theadditional element u∈ S in Definition 6.2 (b) this would be false,since y ·1−0 ·1 6= 0 ∈ A(X). However, if we can set u = x ∈ S,then x(y ·1−0 ·1) = xy = 0 ∈ A(X), and hence y

1 = 01 ∈ S−1R as

desired.

X

a

Remark 6.4. Let S be a multiplicatively closed subset of a ring R.

54 Andreas Gathmann

(a) There is an obvious ring homomorphism ϕ : R→ S−1R, a 7→ a1 that makes S−1R into an

R-algebra. However, ϕ is only injective if S does not contain zero-divisors, as a1 = 0

1 bydefinition only implies the existence of an element u ∈ S with u(a · 1− 0 · 1) = ua = 0. Wehave already seen a concrete example of this in Remark 6.3 (b): in that case we had y 6= 0∈R,but y

1 = 01 ∈ S−1R.

(b) In Definition 6.2 (a) of a multiplicatively closed subset we have not excluded the case 0 ∈ S,i. e. that we “want to allow division by 0”. However, in this case we trivially get S−1R = 0,since we can then always take u = 0 in Definition 6.2 (b).

Example 6.5 (Standard examples of localization). Let us now give some examples of multiplica-tively closed sets in a ring R, and thus of localizations. In fact, the following are certainly the mostimportant examples — probably any localization that you will meet is of one of these forms.

(a) Of course, S= {1} is a multiplicatively closed subset, and leads to the localization S−1R∼=R.

(b) Let S be the set of non-zero-divisors in R. Then S is multiplicatively closed: if a,b ∈ S thenab is also not a zero-divisor, since abc = 0 for some c ∈ R first implies bc = 0 since a ∈ S,and then c = 0 since b ∈ S.

Of particular importance is the case when R is an integral domain. Then S = R\{0}, andevery non-zero element a

s is a unit in S−1R, with inverse sa . Hence S−1R is a field, the

so-called quotient field QuotR of R.

Moreover, in this case every localization T−1R at a multiplicatively closed subset T with0 /∈ T is naturally a subring of QuotR, since

T−1R→ QuotR,as7→ a

sis an injective ring homomorphism. In particular, by (a) the ring R itself can be thought ofas a subring of its quotient field QuotR as well.

(c) For a fixed element a ∈ R let S = {an : n ∈ N}. Then S is obviously multiplicatively closed.The corresponding localization S−1R is often written as Ra; we call it the localization of Rat the element a.

(d) Let PER be a prime ideal. Then S = R\P is multiplicatively closed, since a /∈ P and b /∈ Pimplies ab /∈ P by Definition 2.1 (a). The resulting localization S−1R is usually denoted byRP and called the localization of R at the prime ideal P.

In fact, this construction of localizing a ring at a prime ideal is probably the most importantcase of localization. In particular, if R = A(X) is the ring of functions on a variety X andP = I(a) = { f ∈ A(X) : f (a) = 0} the ideal of a point a ∈ X (which is maximal and henceprime, see Remark 2.7), the localization RP is exactly the ring of local functions on X at aas in Remark 6.1 (b). So localizing a coordinate ring at the maximal ideal corresponding toa point can be thought of as studying the variety locally around this point.

Example 6.6 (Localizations of Z). As concrete examples, let us apply the constructions of Example6.5 to the ring R = Z of integers. Of course, localizing at Z\{0} as in (b) gives the quotient fieldQuotZ = Q as in Remark 6.1 (a). If p ∈ Z is a prime number then localization at the element p asin (c) gives the ring

Zp =

{apn : a ∈ Z,n ∈ N

}⊂Q,

whereas localizing at the prime ideal (p) as in (d) yields

Z(p) =

{ab

: a,b ∈ Z, p 6 | b}⊂Q

(these are both subrings of Q by Example 6.5 (b)). So note that we have to be careful about thenotations, in particular since Zp is usually also used to denote the quotient Z/pZ.

6. Localization 55

After having seen many examples of localizations, let us now turn to the study of their properties.We will start by relating ideals in a localization S−1R to ideals in R.

Proposition 6.7 (Ideals in localizations). Let S be a multiplicatively closed subset of a ring R. In thefollowing, we will consider contractions and extensions by the ring homomorphism ϕ : R→ S−1R.

(a) For any ideal IER we have Ie = { as : a ∈ I,s ∈ S}.

(b) For any ideal IES−1R we have (Ic)e = I.

(c) Contraction and extension by ϕ give a one-to-one correspondence

{prime ideals in S−1R} 1:1←→ {prime ideals I in R with I∩S = /0}

I 7−→ Ic

Ie ←−7 I.

Proof.

(a) As 1s ∈ S−1R for s ∈ S, the ideal Ie generated by ϕ(I) must contain the elements 1

s ·ϕ(a) =as

for a ∈ I. But it is easy to check that { as : a ∈ I,s ∈ S} is already an ideal in S−1R, and so it

has to be equal to Ie.

(b) By Exercise 1.19 (b) it suffices to show the inclusion “⊃”. So let as ∈ I. Then a∈ϕ−1(I) = Ic

since ϕ(a) = a1 = s · a

s ∈ I. By (a) applied to Ic it follows that as ∈ (Ic)e.

(c) We have to check a couple of assertions.

The map “→” is well-defined: If I is a prime ideal in S−1R then Ic is a prime ideal in R byExercise 2.9 (b). Moreover, we must have Ic∩S = /0 as all elements of S are mapped to unitsby ϕ , and I cannot contain any units.

The map “←” is well-defined: Let IER be a prime ideal with I∩S = /0. We have to checkthat Ie is a prime ideal. So let a

s ,bt ∈ S−1R with a

s ·bt ∈ Ie. By (a) this means that ab

st = cu for

some c ∈ I and u ∈ S, i. e. there is an element v ∈ S with v(abu− stc) = 0. This implies thatuvab ∈ I. Since I is prime, one of these four factors must lie in I. But u and v are in S andthus not in I. Hence we conclude that a ∈ I or b ∈ I, and therefore a

s ∈ Ie or bt ∈ Ie by (a).

(Ic)e = I for any prime ideal I in S−1R: this follows from (b).

(Ie)c = I for any prime ideal I in R with I ∩ S = /0: By Exercise 1.19 (a) we only have tocheck the inclusion “⊂”. So let a ∈ (Ie)c, i. e. ϕ(a) = a

1 ∈ Ie. By (a) this means a1 = b

s forsome b ∈ I and s ∈ S, and thus there is an element u ∈ S with u(as− b) = 0. Thereforesua ∈ I, which means that one of these factors is in I since I is prime. But s and u lie in Sand hence not in I. So we conclude that a ∈ I. �

Example 6.8 (Prime ideals in RP). Applying Proposition 6.7 (c) to the case of a localization of aring R at a prime ideal P, i. e. at the multiplicatively closed subset R\P as in Example 6.5 (d), givesa one-to-one correspondence by contraction and extension

{prime ideals in RP}1:1←→ {prime ideals I in R with I ⊂ P}. (∗)

One should compare this to the case of prime ideals in the quotient ring R/P: by Lemma 1.21 andCorollary 2.4 we have a one-to-one correspondence

{prime ideals in R/P} 1:1←→ {prime ideals I in R with I ⊃ P},this time by contraction and extension by the quotient map. Unfortunately, general ideals do notbehave as nicely: whereas ideals in the quotient R/P still correspond to ideals in R containing P, theanalogous statement for general ideals in the localization RP would be false.

As we will see now, another consequence of the one-to-one correspondence (∗) is that the localiza-tion RP has exactly one maximal ideal (of course it always has at least one by Corollary 2.17). Sincethis is a very important property of rings, it has a special name.

56 Andreas Gathmann

Definition 6.9 (Local rings). A ring is called local if it has exactly one maximal ideal.

Corollary 6.10 (RP is local). Let P be a prime ideal in a ring R. Then RP is local with maximalideal Pe = { a

s : a ∈ P,s /∈ P}.

Proof. Obviously, the set of all prime ideals of R contained in P has the upper bound P. So asthe correspondence (∗) of Example 6.8 preserves inclusions, we see that every prime ideal in RP iscontained in Pe. In particular, every maximal ideal in RP must be contained in Pe. But this meansthat Pe is the only maximal ideal. The formula for Pe is just Proposition 6.7 (a). �

Note however that arbitrary localizations as in Definition 6.2 are in general not local rings.

Remark 6.11 (Rings of local functions on a variety are local). If R = A(X) is the coordinate ringof a variety X and P = I(a) the maximal ideal of a point a ∈ X , the localization RP is the ring oflocal functions on X at a as in Example 6.5 (d). By Corollary 6.10, this is a local ring with uniquemaximal ideal Pe, i. e. with only the maximal ideal corresponding to the point a. This fits nicelywith the interpretation of Remark 6.1 (b) that RP describes an “arbitrarily small neighborhood” of a:although every function in RP is defined on X in a neighborhood of a, there is no point except a atwhich every function in RP is well-defined.

Exercise 6.12 (Universal property of localization). Let S be a multiplicatively closed subset of aring R. Prove that the localization S−1R satisfies the following universal property: for any homomor-phism α : R→ R′ to another ring R′ that maps S to units in R′ there is a unique ring homomorphismϕ : S−1R→ R′ such that ϕ( r

1 ) = α(r) for all r ∈ R.

Exercise 6.13.

(a) Let R = R[x,y]/(xy) and P = (x− 1)ER. Show that P is a maximal ideal of R, and thatRP ∼= R[x](x−1). What does this mean geometrically?

(b) Let R be a ring and a ∈ R. Show that Ra ∼= R[x]/(ax−1), where Ra denotes the localizationof R at a as in Example 6.5 (c). Can you give a geometric interpretation of this statement?

Exercise 6.14. Let S be a multiplicatively closed subset in a ring R, and let IER be an ideal withI∩S = /0. Prove the following “localized version” of Corollary 2.17:

(a) The ideal I is contained in a prime ideal P of R such that P∩S = /0 and S−1P is a maximalideal in S−1R.

(b) The ideal I is not necessarily contained in a maximal ideal P of R with P∩S = /0.

Exercise 6.15 (Saturations). For a multiplicatively closed subset S of a ring R we call

S := {s ∈ R : as ∈ S for some a ∈ R}

the saturation of S. Show that S is a multiplicatively closed subset as well, and that S−1R∼= S−1R.

Exercise 6.16 (Alternative forms of Nakayama’s Lemma for local rings). Let M be a finitely gen-erated module over a local ring R. Prove the following two versions of Nakayama’s Lemma (seeCorollary 3.27):

(a) If there is a proper ideal IER with IM = M, then M = 0.

(b) If IER is the unique maximal ideal and m1, . . . ,mk ∈ M are such that m1, . . . ,mk generateM/IM as an R/I-vector space, then m1, . . . ,mk generate M as an R-module.

So far we have only applied the technique of localization to rings. But in fact this works equallywell for modules, so let us now give the corresponding definitions. However, as the computations tocheck that these definitions make sense are literally the same as in Lemma 6.2, we will not repeatthem here.

6. Localization 57

Definition 6.17 (Localization of modules). Let S be a multiplicatively closed subset of a ring R, andlet M be an R-module. Then

(m,s)∼ (m′,s′) :⇔ there is an element u ∈ S such that u(s′m− sm′) = 0

is an equivalence relation on M× S. We denote the equivalence class of (m,s) ∈M× S by ms . The

set of all equivalence classes

S−1M :={m

s: m ∈M,s ∈ S

}is then called the localization of M at S. It is an S−1R-module together with the addition and scalarmultiplication

ms+

m′

s′:=

s′m+ sm′

ss′and

as· m′

s′:=

am′

ss′

for all a ∈ R, m,m′ ∈M, and s,s′ ∈ S.

In the cases of example 6.5 (c) and (d), when S is equal to {an : n ∈ N} for an element a ∈ R or R\Pfor a prime ideal PER, we will write S−1M also as Ma or MP, respectively.

Example 6.18. Let S be a multiplicatively closed subset of a ring R, and let I be an ideal in R. Thenby Proposition 6.7 (a) the localization S−1I in the sense of Definition 6.17 is just the extension Ie ofI with respect to the canonical map R→ S−1R, a 7→ a

1 .10

In fact, it turns out that the localization S−1M of a module M is just a special case of a tensor product.As one might already expect from Definition 6.17, it can be obtained from M by an extension ofscalars from R to the localization S−1R as in Construction 5.16. Let us prove this first, so that wecan then carry over some of the results of Chapter 5 to our present situation.

Lemma 6.19 (Localization as a tensor product). Let S be a multiplicatively closed subset in a ringR, and let M be an R-module. Then S−1M ∼= M⊗R S−1R.

Proof. There is a well-defined R-module homomorphism

ϕ : S−1M→M⊗R S−1R,ms7→ m⊗ 1

s,

since for m′ ∈M and s′ ∈ S with ms = m′

s′ , i. e. such that u(s′m− sm′) = 0 for some u ∈ S, we have

m⊗ 1s−m′⊗ 1

s′= u(s′m− sm′)⊗ 1

uss′= 0.

Similarly, by the universal property of the tensor product we get a well-defined homomorphism

ψ : M⊗R S−1R→ S−1M with ψ

(m⊗ a

s

)=

ams,

as for a′ ∈ R and s′ ∈ S with u(s′a−sa′) = 0 for some u∈ S we also get u(s′am−sa′m) = 0, and thusams = a′m

s′ . By construction, ϕ and ψ are inverse to each other, and thus ϕ is an isomorphism. �

Remark 6.20 (Localization of homomorphisms). Let ϕ : M→N be a homomorphism of R-modules,and let S ⊂ R be a multiplicatively closed subset. Then by Construction 5.16 there is a well-definedhomomorphism of S−1R-modules

ϕ⊗ id : M⊗R S−1R→ N⊗R S−1R such that (ϕ⊗ id)(

m⊗ as

)= ϕ(m)⊗ a

s.

By Lemma 6.19 this can now be considered as an S−1R-module homomorphism

S−1ϕ : S−1M→ S−1N with S−1

ϕ

(ms

)=

ϕ(m)

s.

We will call this the localization S−1ϕ of the homomorphism ϕ . As before, we will denote it by ϕaor ϕP if S = {an : n ∈ N} for an element a ∈ R or S = R\P for a prime ideal PER, respectively.

58 Andreas Gathmann

The localization of rings and modules is a very well-behaved concept in the sense that it is compatiblewith almost any other construction that we have seen so far. One of the most important propertiesis that it preserves exact sequences, which then means that it is also compatible with everything thatcan be expressed in terms of such sequences.

Proposition 6.21 (Localization is exact). For every short exact sequence

0−→M1ϕ−→M2

ψ−→M3 −→ 0

of R-modules, and any multiplicatively closed subset S⊂ R, the localized sequence

0−→ S−1M1S−1ϕ−→ S−1M2

S−1ψ−→ S−1M3 −→ 0

is also exact.

Proof. By Lemma 6.19, localization at S is just the same as tensoring with S−1R. But tensor productsare right exact by Proposition 5.22 (b), hence it only remains to prove the injectivity of S−1ϕ .

So let ms ∈ S−1M1 with S−1ϕ(m

s ) =ϕ(m)

s = 0. This means that there is an element u ∈ S withuϕ(m) = ϕ(um) = 0. But ϕ is injective, and therefore um = 0. Hence m

s = 1us ·um = 0, i. e. S−1ϕ is

injective. �

Corollary 6.22. Let S be a multiplicatively closed subset of a ring R.

(a) For any homomorphism ϕ : M → N of R-modules we have ker(S−1ϕ) = S−1 kerϕ andim(S−1ϕ) = S−1 imϕ .

In particular, if ϕ is injective / surjective then S−1ϕ is injective / surjective, and if M is asubmodule of N then S−1M is a submodule of S−1N in a natural way.

(b) If M is a submodule of an R-module N then S−1(N/M)∼= S−1N/S−1M.

(c) For any two R-modules M and N we have S−1(M⊕N)∼= S−1M⊕S−1N.

Proof.

(a) Localizing the exact sequence 0 −→ kerϕ −→Mϕ−→ imϕ −→ 0 of Example 4.3 (a) at S,

we get by Proposition 6.21 an exact sequence

0−→ S−1 kerϕ −→ S−1MS−1ϕ−→ S−1 imϕ −→ 0.

But again by Example 4.3 this means that the modules S−1 kerϕ and S−1 imϕ in this se-quence are the kernel and image of the map S−1ϕ , respectively.

(b) This time we localize the exact sequence 0 −→ M −→ N −→ N/M −→ 0 of Example 4.3(b) to obtain by Proposition 6.21

0−→ S−1M −→ S−1N −→ S−1(N/M)→ 0,

which by Example 4.3 means that the last module S−1(N/M) in this sequence is isomorphicto the quotient S−1N/S−1M of the first two.

(c) This follows directly from the definition, or alternatively from Lemma 5.8 (c) as localizationis the same as taking the tensor product with S−1R. �

Remark 6.23. If an operation such as localization preserves short exact sequences as in Proposition6.21, the same then also holds for longer exact sequences: in fact, we can split a long exact sequenceof R-modules into short ones as in Remark 4.5, localize it to obtain corresponding short exact se-quences of S−1R-modules, and then glue them back to a localized long exact sequence by Lemma4.4 (b).

Exercise 6.24. Let S be a multiplicatively closed subset of a ring R, and let M be an R-module.Prove:

(a) For M1,M2 ≤M we have S−1(M1 +M2) = S−1M1 +S−1M2.

6. Localization 59

(b) For M1,M2 ≤M we have S−1(M1∩M2) = S−1M1∩S−1M2. However, if Mi ≤M for all i inan arbitrary index set J, it is in general not true that S−1

(⋂i∈J Mi

)=⋂

i∈J S−1Mi. Does atleast one inclusion hold?

(c) For an ideal IER we have S−1√

I =√

S−1I.

Example 6.25. Let P and Q be maximal ideals in a ring R; we want to compute the ring RQ/PQ.Note that this is also an RQ-module, and as such isomorphic to (R/P)Q by Corollary 6.22 (b). Sowhen viewed as such a module, it does not matter whether we first localize at Q and then take thequotient by P or vice versa.

(a) If P 6= Q we must also have P 6⊂ Q since P is maximal and Q 6= R. So there is an elementa ∈ P\Q, which leads to 1

1 = aa ∈ PQ. Hence in this case the ideal PQ in RQ is the whole ring,

and we obtain RQ/PQ = (R/P)Q = 0.

(b) On the other hand, if P = Q there is a well-defined ring homomorphism

ϕ : R/P→ RP/PP, a 7→( a

1

).

We can easily construct an inverse of this map as follows: the morphism R→ R/P, a 7→ asends R\P to units since R/P is a field by Lemma 2.3 (b), hence it passes to a morphism

RP→ R/P,as7→ s−1 a

by Exercise 6.12. But PP is in the kernel of this map, and thus we get a well-defined ringhomomorphism

RP/PP→ R/P,( a

s

)7→ s−1 a,

which clearly is an inverse of ϕ . So the ring RP/PP is isomorphic to R/P.

Note that this result is also easy to understand from a geometric point of view: let R be the ringof functions on a variety X over an algebraically closed field, and let p,q ∈ X be the points corre-sponding to the maximal ideals P and Q, respectively (see Remark 2.7 (a)). Then taking the quotientby P = I(p) means restricting the functions to the point p by Lemma 0.9 (d), and localizing at Qcorresponds to restricting them to an arbitrarily small neighborhood of q by Example 6.5 (d).

So first of all we see that the order of these two operations should not matter, since restrictions offunctions commute. Moreover, if p 6= q then our small neighborhood of q does not meet p, and weget the zero ring as the space of functions on the empty set. In contrast, if p = q then after restrictingthe functions to the point p another restriction to a neighborhood of p does not change anything, andthus we obtain the isomorphism of (b) above.

As an application of this result, we get the following refinement of the notion of length of a module:not only is the length of all composition series the same, but also the maximal ideals occurring inthe quotients of successive submodules in the composition series as in Definition 3.18 (a).

Corollary 6.26 (Length of localized modules). Let M be an R-module of finite length, and let

0 = M0 ( M1 ( · · ·( Mn = M

be a composition series for M, so that Mi/Mi−1 ∼= R/Ii by Definition 3.18 (a) with some maximalideals IiER for i = 1, . . . ,n.

Then for any maximal ideal PER the localization MP is of finite length as an RP-module, and itslength lRP(MP) is equal to the number of indices i such that Ii = P. In particular, up to permutationsthe ideals Ii (and their multiplicities) as above are the same for all choices of composition series.

Proof. Let PE R be a maximal ideal. From the given composition series we obtain a chain oflocalized submodules

0 = (M0)P ⊂ (M1)P ⊂ ·· · ⊂ (Mn)P = MP. (∗)

60 Andreas Gathmann

For its successive quotients (Mi)P/(Mi−1)P for i = 1, . . . ,n we get by Corollary 6.22 (b)

(Mi)P/(Mi−1)P ∼= (Mi/Mi−1)P ∼= (R/Ii)P ∼= RP/(Ii)P.

But by Example 6.25 this is 0 if Ii 6= P, whereas for Ii = P it is equal to RP modulo its uniquemaximal ideal PP by Corollary 6.10. So by deleting all equal submodules from the chain (∗) weobtain a composition series for MP over RP whose length equals the number of indices i ∈ {1, . . . ,n}with Ii = P. �

We can rephrase Corollary 6.26 by saying that the length of a module is a “local concept”: if anR-module M is of finite length and we know the lengths of the localized modules MP for all maximalideals PER, we also know the length of M: it is just the sum of all the lengths of the localizedmodules.

In a similar way, there are many other properties of an R-module M that carry over from the localizedmodules MP, i. e. in geometric terms from considering M locally around small neighborhoods ofevery point. This should be compared to properties in geometric analysis such as continuity anddifferentiability: if we have e. g. a function f : U → R on an open subset U of R we can definewhat it means that it is continuous at a point p ∈U — and to check this we only need to know fin an arbitrarily small neighborhood of p. In total, the function f is then continuous if and only ifit is continuous at all points p ∈U : we can say that “continuity is a local property”. Here are somealgebraic analogues of such local properties:

Proposition 6.27 (Local properties). Let R be a ring.

(a) If M is an R-module with MP = 0 for all maximal ideals PER, then M = 0.

(b) Consider a sequence of R-module homomorphisms

0−→M1ϕ1−→M2

ϕ2−→M3 −→ 0. (∗)

If the localized sequences of RP-modules

0−→ (M1)P(ϕ1)P−→ (M2)P

(ϕ2)P−→ (M3)P −→ 0

are exact for all maximal ideals PER, then the original sequence (∗) is exact as well.

We say that being zero and being exact are “local properties” of a module and a sequence, respec-tively. Note that the converse of both statements holds as well by Proposition 6.21.

Proof.

(a) Assume that M 6= 0. If we then choose an element m 6= 0 in M, its annihilator ann(m) doesnot contain 1 ∈ R, so it is a proper ideal of R and thus by Corollary 2.17 contained in amaximal ideal P. Hence for all u ∈ R\P we have u /∈ ann(m), i. e. um 6= 0 in M. This meansthat m

1 is non-zero in MP, and thus MP 6= 0.

(b) It suffices to show that a three-term sequence Lϕ−→ M

ψ−→ N is exact if all localizationsLP

ϕP−→ MPψP−→ NP are exact for maximal PER, since we can then apply this to all three

possible positions in the five-term sequence of the statement.

To prove this, note first that

(ψ(ϕ(L)))P =

{ψ(ϕ(m))

s: m ∈ L,s ∈ S

}={

ψP

(ϕP

(ms

)): m ∈ L,s ∈ S

}= ψP(ϕP(LP)) = 0,

since the localized sequence is exact. This means by (a) that ψ(ϕ(L)) = 0, i. e. imϕ ⊂ kerψ .We can therefore form the quotient kerψ/ imϕ , and get by the exactness of the localizedsequence

(kerψ/ imϕ)P ∼= (kerψ)P/(imϕ)P = kerψP/ imϕP = 0

6. Localization 61

using Corollary 6.22. But this means kerψ/ imϕ = 0 by (a), and so kerψ = imϕ , i. e. thesequence L

ϕ−→Mψ−→ N is exact. �

Remark 6.28. As usual, Proposition 6.27 (b) means that all properties that can be expressed in termsof exact sequences are local as well: e. g. an R-module homomorphism ϕ : M→ N is injective (resp.surjective) if and only if its localizations ϕP : MP → NP for all maximal ideals PER are injective(resp. surjective). Moreover, as in Remark 6.23 it follows that Proposition 6.27 (b) holds for longerexact sequences as well.

Exercise 6.29. Let R be a ring. Show:

(a) Ideal containment is a local property: for two ideals I,JER we have I ⊂ J if and only ifIP ⊂ JP in RP for all maximal ideals PER.

(b) Being reduced is a local property, i. e. R is reduced if and only if RP is reduced for allmaximal ideals PER.

(c) Being an integral domain is not a local property, i. e. R might not be an integral domainalthough the localizations RP are integral domains for all maximal ideals PER. Can yougive a geometric interpretation of this statement?

62 Andreas Gathmann

7. Chain Conditions

In the previous chapters we have met several finiteness conditions: an algebra can be finitely gener-ated as in Definition 1.26 (b), a module can also be finitely generated as in Definition 3.3 (b) or havefinite length as in Definition 3.18, and in our study of Zorn’s Lemma in Remark 2.13 we discussedwhether an increasing chain of ideals in a ring has to stop after finitely many steps. Of course, thiscan often make a difference: for example, a reduced algebra over a field is the coordinate ring ofa variety if and only if it is finitely generated (see Remark 1.31), and the Cayley-Hamilton theo-rem in Proposition 3.25 together with its corollaries such as Nakayama’s Lemma in Corollary 3.27only hold for finitely generated modules. So we will now take some time to study such finitenessquestions in more detail and see how they are related.

The key idea is to consider chains of submodules in a given module and check whether they have tostop after finitely many terms.

Definition 7.1 (Noetherian and Artinian modules). Let M be an R-module.

(a) M is called Noetherian if every ascending chain

M0 ⊂M1 ⊂M2 ⊂ ·· ·of submodules of M becomes stationary, i. e. if for every such chain there is an index n ∈ Nsuch that Mk = Mn for all k≥ n. Obviously, this is the same as saying that there is no infinitestrictly ascending chain M0 ( M1 ( M2 ( · · · .

(b) Similarly, M is called Artinian if every descending chain

M0 ⊃M1 ⊃M2 ⊃ ·· ·of submodules becomes stationary. Again, this is equivalent to requiring that there is noinfinite strictly descending chain M0 ) M1 ) M2 ) · · · .

The conditions of (a) and (b) are often referred to as the ascending and descending chain condition,respectively. The ring R itself is called Noetherian or Artinian if it has this property as an R-module;the submodules above are then just ideals of R by Example 3.4 (a).

Example 7.2.(a) Any field K is trivially Noetherian and Artinian as it has only the trivial ideals (0) and K.

More generally, a K-vector space V is Noetherian if and only if it is Artinian if and only if itis finite-dimensional (i. e. finitely generated):

• If V is finite-dimensional, there can only be finite strictly ascending or descendingchains of vector subspaces of V since the dimension has to be strictly increasing ordecreasing in such a chain, respectively.

• If V is infinite-dimensional, we can obviously form a chain M0 ( M1 ( M2 ( · · · withdimK Mn = n for all n∈N: set M0 = 0, and Mn+1 =Mn+〈vn+1 〉with vn+1 /∈Mn for alln ∈ N. Similarly, we can also find an infinite descending chain M0 ) M1 ) M2 ) · · ·of infinite-dimensional subspaces of V with dimK(M/Mn) = n for all n: let M0 = M,and Mn+1 = Mn ∩ kerϕn+1 for some linear map ϕn+1 : V → K that is not identicallyzero on Mn. Then M/Mn ∼= (M/Mn+1)/(Mn/Mn+1) by Proposition 3.10 (b), and sodimMn/Mn+1 = 1 implies dimM/Mn = n for all n by induction.

(b) The ring Z is Noetherian: if we had a strictly increasing chain of ideals I0 ( I1 ( I2 ( · · ·in Z, then certainly I1 6= 0, and thus I1 = (a) for some non-zero a ∈ Z. But there are onlyfinitely many ideals in Z that contain I1 since they correspond to ideals of the finite ringZ/(a) by Lemma 1.21. Hence the chain cannot be infinitely long, and thus Z is Noetherian.

7. Chain Conditions 63

On the other hand, Z is not Artinian, since there is an infinite decreasing chain of ideals

Z ) 2Z ) 4Z ) 8Z ) · · · .

(c) Let R =⋃

n∈NR[x0,x1, . . . ,xn] be the polynomial ring over R in infinitely many variables.Then R is neither Noetherian nor Artinian, since there are infinite chains of ideals

(x0)( (x0,x1)( (x0,x1,x2)( · · · and (x0)) (x20)) (x3

0)) · · · .

Exercise 7.3. For a prime number p ∈ N, consider M = Zp/Z as a Z-module (i. e. as an Abeliangroup), where Zp ⊂ Q denotes the localization of Z at the element p as in Example 6.5 (c). Showthat:

(a) Every proper submodule of M is of the form⟨

1pn

⟩.

(b) M is Artinian, but not Noetherian.

As you might expect, we will see in the following that Noetherian and Artinian modules have manysimilar properties that can be obtained from one another by just reversing all inclusions — as e. g. inthe first two parts of the following lemma. However, there are also aspects in which the two conceptsof Noetherian and Artinian modules are fundamentally different (in particular when we specialize torings in the second half of this chapter). A first example of this is part (c) of the following equivalentreformulation of our chain conditions, which asserts that a module is Noetherian if and only if eachof its submodules is finitely generated. There is no corresponding condition for Artinian modules,and in fact this is one of the main reasons why in practice Noetherian modules are much moreimportant than Artinian ones.

Lemma 7.4 (Equivalent conditions for Noetherian and Artinian modules). Let M be an R-module.

(a) M is Noetherian if and only if every non-empty family of submodules of M has a maximalelement.

(b) M is Artinian if and only if every non-empty family of submodules of M has a minimalelement.

(c) M is Noetherian if and only if every submodule of M is finitely generated.

Proof.

(a) “⇒” Let A be a non-empty family of submodules of M. If there was no maximal elementof A, we could choose recursively a chain M0 ( M1 ( M2 ( · · · from A, contradictingthe assumption that M is Noetherian.

“⇐” Consider a chain M0 ⊂M1 ⊂M2 ⊂ ·· · of submodules of M. By assumption, the setA = {Mn : n ∈ N} has a maximal element Mn. But then Mk = Mn for all k ≥ n, henceM is Noetherian.

(b) is proven in the same way as (a), just reversing all inclusions. 11

(c) “⇒” Assume that we have a submodule N ≤M that is not finitely generated. Then we canrecursively pick m0 ∈ N and mn+1 ∈ N\〈m0, . . . ,mn 〉 for n ∈ N, and obtain a chainM0 ( M1 ( M2 ( · · · in M. This is a contradiction since M is Noetherian.

“⇐” Let M0 ⊂ M1 ⊂ M2 ⊂ ·· · be a chain of submodules of M. Then N =⋃

n∈N Mn isalso a submodule of M (which can be shown in the same way as in the proof ofCorollary 2.17). So by assumption N can be generated by finitely many elementsm1, . . . ,mr ∈ N. Now by definition of N we must have mi ∈Mni for all i = 1, . . . ,r andsome n1, . . . ,nr ∈ N. With n = max{n1, . . . ,nr} we then have m1, . . . ,mr ∈Mn. HenceN = 〈m1, . . . ,mr 〉 ≤Mn ≤ N, which implies Mk = Mn = N for all k ≥ n. �

Example 7.5.(a) Any principal ideal domain R (as e. g. R = Z in Example 7.2 (b)) is Noetherian by Lemma

7.4 (c), since every ideal in R is even generated by one element.

64 Andreas Gathmann

(b) Note that the R-algebra R[x] is a Noetherian ring by (a), but not a Noetherian R-moduleby Example 7.2 (a) (as it is an infinite-dimensional R-vector space). So when applying thechain conditions to algebras we have to be very careful whether we talk about Noetherianresp. Artinian rings or modules, as this might make a difference! There is one importantcase however in which there is no such difference:

(c) Let I be an ideal in a ring R, and let M be an R-module. Then M/IM is both an R/I-moduleand an R-module, and by definition a subset of M/IM is an R/I-submodule if and only if it isan R-submodule. So we conclude by Definition 7.1 that M/IM is Noetherian resp. Artinianas an R/I-module if and only if it has this property as an R-module.

In particular, applying this result to M = R we see that the R-algebra R/I is Noetherian resp.Artinian as a ring if and only if it has this property as an R-module.

Remark 7.6 (Maximal ideals in Noetherian rings). Let I be an ideal in a Noetherian ring R withI 6= R. Then every ascending chain of ideals in R becomes stationary, so by Remark 2.13 this meansthat the existence of a maximal ideal in R that contains I is trivial and does not require Zorn’s Lemma(for which we had to work quite a bit). In fact, this is just the statement of Lemma 7.4 (a) which tellsus that the family of all proper ideals of R containing I must have a maximal element.

Let us now prove some basic properties of Noetherian and Artinian modules.

Lemma 7.7. Let N be a submodule of an R-module M.

(a) M is Noetherian if and only if N and M/N are Noetherian.

(b) M is Artinian if and only if N and M/N are Artinian.

Proof. We just prove (a), since (b) can be proven in the same way, reversing all inclusions.

“⇒” Let M be Noetherian. First of all, any chain N0 ⊂ N1 ⊂ N2 ⊂ ·· · of submodules of N is alsoa chain of submodules of M, and thus becomes stationary. Hence N is Noetherian.

Similarly, let P0 ⊂ P1 ⊂ P2 ⊂ ·· · be a chain of submodules of M/N. If we set Mk = q−1(Pk)for all k ∈ N, where q : M→M/N is the quotient map, then M0 ⊂M1 ⊂M2 ⊂ ·· · is a chainof submodules of M. As M is Noetherian, we have Mk = Mn for all k ≥ n with some n ∈ N.But since q is surjective we then also have Pk = q(Mk) = q(Mn) = Pn for all k ≥ n. HenceM/N is Noetherian.

“⇐” Let M0 ⊂M1 ⊂M2 ⊂ ·· · be an ascending chain of submodules in M. If we set Nk := Mk∩Nand Pk := (Mk +N)/N for all k ∈ N, then

N0 ⊂ N1 ⊂ N2 ⊂ ·· · and P0 ⊂ P1 ⊂ P2 ⊂ ·· ·are chains of submodules of N and M/N, respectively. By assumption, both of them becomestationary, and hence there is an element n ∈ N such that Nk = Nn and Pk = Pn for all k ≥ n.But then we obtain a commutative diagram for all k ≥ n

0 Nn Mn Pn 0

0 Nk Mk Pk 0

whose rows are exact by Proposition 3.10 (c) and Example 4.3 (b), and whose columns areinduced by the inclusions Mn→Mk. As the left and right vertical map are isomorphisms, sois the middle one by Corollary 4.12, and thus we have Mk = Mn for k ≥ n as well. Hence Mis Noetherian. �

Remark 7.8.(a) Of course, by Example 4.3 we can rephrase Lemma 7.7 by saying that for any short exact

sequence 0 −→ N −→M −→ P −→ 0 of R-modules the middle entry M is Noetherian (orArtinian) if and only if N and P are Noetherian (or Artinian).


(b) Let I be an ideal in a Noetherian (resp. Artinian) ring R. Combining Lemma 7.7 with Exam-ple 7.5 (c) we see that then the quotient R/I is a Noetherian (resp. Artinian) ring as well.

Corollary 7.9. Let M and N be R-modules.

(a) The direct sum M⊕N is Noetherian if and only if M and N are Noetherian.

(b) If R is Noetherian and M is finitely generated, then M is also Noetherian.

The same statements also hold with “Noetherian” replaced by “Artinian”.

Proof. Again, we only show the statement for Noetherian modules, since the Artinian counterpartfollows in exactly the same way.

(a) By Remark 7.8 (a), this follows from the exact sequence 0−→M −→M⊕N −→ N −→ 0.

(b) Let M = 〈m1, . . . ,mk 〉 for some m1, . . . ,mk ∈M. Then the ring homomorphism

ϕ : Rk→M, (a1, . . . ,ak) 7→ a1m1 + · · ·+akmk

is surjective, so that we have an exact sequence 0−→ kerϕ −→ Rk ϕ−→M −→ 0. Now as Ris Noetherian, so is Rk by (a), and hence also M by Remark 7.8 (a). �

Remark 7.10 (Structure Theorem for finitely generated Abelian groups). Let M be a finitely gener-ated Abelian group, viewed as a finitely generated Z-module as in Example 3.2 (d). Of course, Mis then a Noetherian Z-module by Corollary 7.9 (b). But we can follow the idea of the proof of thisstatement one step further: the Z-module kerϕ occurring in the proof is (again by Remark 7.8 (a))finitely generated as well, and so we also find a surjective ring homomorphism ψ : Zl → kerϕ forsome l ∈ N. This results in another short exact sequence 0 −→ kerψ −→ Zl ψ−→ kerϕ −→ 0, andthus by gluing as in Lemma 4.4 (b) in the exact sequence

0−→ kerψ −→ Zl ψ−→ Zk ϕ−→M −→ 0,

which means that M ∼= Zk/kerϕ = Zk/ imψ .

Now ψ ∈HomZ(Zl ,Zk) is just given by an integer k× l matrix by Remark 3.17 (b). Similarly to thecase of matrices over a field [G2, Proposition 16.43] one can show that it is possible to change basesin Zl and Zk such that the matrix of ψ has non-zero entries only on the diagonal. But this means thatimψ is generated by a1e1, . . . ,akek for some a1, . . . ,ak ∈ Z, where e1, . . . ,ek denotes the standardbasis of Zk. Thus

M ∼= Zk/ imψ ∼= Z/a1Z×·· ·×Z/akZ,and so we conclude that every finitely generated Abelian group is a product of cyclic groups. Ofcourse, by the Chinese Remainder Theorem [G1, Proposition 11.22] this can also be rewritten as

M ∼= Zr×Z/q1Z×·· ·×Z/qnZfor r,n ∈ N and (not necessarily distinct) prime powers q1, . . . ,qn.

Let us now see how the finite length condition on modules is related to the concepts introduced inthis chapter.

Lemma 7.11. An R-module M is of finite length if and only if it is both Noetherian and Artinian.

Proof. If M is of finite length, then all strict chains of submodules of M are finite by Exercise 3.19(b) and (c). So in this case M is clearly both Noetherian and Artinian.

Conversely, assume that M is both Noetherian and Artinian. Starting from M0 = 0, we try to constructa chain M0 ( M1 ( M2 ( · · · of submodules of M as follows: for n ∈ N let Mn+1 be a minimalsubmodule of M that strictly contains Mn — as long as Mn 6= M this works by Lemma 7.4 (b) sinceM is Artinian. But as M is Noetherian as well, we cannot get such an infinite ascending chain ofsubmodules, and thus we conclude that we must have Mn = M for some n ∈ N. The resulting chain0 = M0 ( M1 ( · · ·( Mn = M is then a composition series for M, since by construction there are nosubmodules between Mi−1 and Mi for all i = 1, . . . ,n. �

66 Andreas Gathmann

Exercise 7.12. Let M be an R-module, and let ϕ : M→M be an R-module homomorphism. If M isNoetherian (hence finitely generated) and ϕ is surjective, you already know by Corollary 3.28 thatϕ has to be an isomorphism.

Now show that if M is Artinian and ϕ is injective, then ϕ is again an isomorphism.

(Hint: Consider the images of ϕn for n ∈ N.)

So far we have mostly considered chain conditions for general modules. For the rest of this chapterwe now want to specialize to the case of rings. In this case we can obtain stronger results, howeverwe will also see that this is where the Noetherian and Artinian conditions begin to diverge drastically.So let us consider these two conditions in turn, starting with the more important case of Noetherianrings.

The one central result on Noetherian rings is Hilbert’s Basis Theorem, which implies that “mostrings that you will meet in practice are Noetherian”.

Proposition 7.13 (Hilbert’s Basis Theorem). If R is a Noetherian ring, then so is the polynomialring R[x].

Proof. Assume that R[x] is not Noetherian. Then by Lemma 7.4 (c) there is an ideal IER[x] that isnot finitely generated. We can therefore pick elements f0, f1, f2, . . . ∈ I as follows: let f0 ∈ I be anon-zero polynomial of minimal degree, and for k ∈ N let fk+1 be a polynomial of minimal degreein I\〈 f0, . . . , fk 〉.Now for all k∈N let dk ∈N be the degree and ak ∈R the leading coefficient of fk, so that we can writefk = ak xdk + (lower order terms). Note that dk ≤ dk+1 for all k by construction of the polynomials.Moreover, since R is Noetherian the chain of ideals (a0) ⊂ (a0,a1) ⊂ (a0,a1,a2) ⊂ ·· · becomesstationary, and thus we must have an+1 = c0a0+ · · ·+cnan for some n∈N and c0, . . . ,cn ∈R. We cantherefore cancel the leading term in fn+1 by subtracting a suitable linear combination of f0, . . . , fn:in the polynomial

f ′n+1 := fn+1−n

∑k=0

ckxdn+1−dk fk

the xdn+1 -coefficient is an+1−c0a0−·· ·−cnan = 0. But this means that deg f ′n+1 < deg fn+1, and asfn+1 /∈ 〈 f0, . . . , fn 〉 we must have f ′n+1 /∈ 〈 f0, . . . , fn 〉 as well. This contradicts our choice of fn+1,proving that an ideal I as above cannot exist, and thus that R[x] is Noetherian. �

Corollary 7.14. Any finitely generated algebra over a Noetherian ring is itself a Noetherian ring.

Proof. Let R be a Noetherian ring. By Lemma 1.30, any finitely generated R-algebra is of theform R[x1, . . . ,xn]/I for an ideal I in the polynomial ring R[x1, . . . ,xn]. Hilbert’s Basis Theorem nowimplies by induction that the polynomial ring R[x1, . . . ,xn] = R[x1][x2] · · · [xn] is Noetherian. So byRemark 7.8 (b) the quotient R[x1, . . . ,xn]/I is a Noetherian ring as well. �

Remark 7.15 (Geometric interpretation of Noetherian and Artinian rings). We have seen in Remark1.31 that any coordinate ring A(X) of a variety X over a field K is a finitely generated K-algebra.So by Corollary 7.14 we see that A(X) is always a Noetherian ring. In particular, by Lemma 7.4 (c)this means that every ideal in A(X) is finitely generated, and hence that any subvariety of X can bedefined by finitely many polynomial equations.

It is also instructive to study the original chain conditions of Definition 7.1 in a geometric setting. Asthe correspondence of Remark 1.10 between (radical) ideals in A(X) and subvarieties of X reversesinclusions, the ascending chain condition on ideals for the Noetherian ring A(X) translates into adescending chain condition on subvarieties of X , i. e. every chain

X0 ⊃ X1 ⊃ X2 ⊃ ·· ·of subvarieties of X must become stationary. The geometric picture behind this is the following:to make a subvariety smaller one has to drop an irreducible component or to reduce the dimensionof the subvariety (a concept that we will introduce in Chapter 11), and this process must always


terminate since the number of components and the dimension are natural numbers. For example, inX = A2

R we could have the following descending chain of subvarieties:

) ) ) ) ) )

R2 V (xy) V (y) V (y,x3− x) V (y,x2− x) V (y,x) /0

Of course, we can easily construct finite descending chains of subvarieties of A2R of any length in

the same way, but infinite chains are impossible.

In contrast, as soon as X contains infinitely many points a1,a2,a3, . . . , it is easy to construct aninfinite strictly ascending chain of subvarieties X0 ( X1 ( X2 ( of X by setting Xn = {ak : k ≤ n}for all n ∈ N. As this corresponds to a strictly decreasing chain of ideals in A(X), we expect that acoordinate ring A(X) is Artinian if and only if X is a finite collection of points — so that the Artiniancondition is a very strong one, in contrast to the Noetherian one.

12To turn these expectations into rigorous statements, let us now study Artinian rings in detail andprove some algebraic results that all correspond to the geometric idea that an Artinian ring R shoulddescribe a finite union of points X . More precisely, consider the correspondence of subvarieties ofX and ideals of R as in Remark 2.7: as the only irreducible subvarieties of X are single points, wewould expect that any prime ideal of R is already maximal. Let us prove this now, together with thefact that in an Artinian ring the zero ideal is always a product of maximal ideals — which can alsobe translated into geometry by Remark 1.12 by saying that X is a union of finitely many points.

Proposition 7.16. Let R be an Artinian ring.

(a) There are (not necessarily distinct) maximal ideals P1, . . . ,PnER such that P1 · · · · ·Pn = 0.

(b) R has only finitely many prime ideals, all of them are maximal, and occur among theP1, . . . ,Pn in (a).

Proof.

(a) Let I = P1 · · · · ·Pn be a product of maximal ideals P1, . . . ,Pn such that I is minimal amongall ideals that can be written in this form — such a minimal element exists by Lemma 7.4(b) since R is Artinian. We need to show that I = 0. First we note:

(i) I2 is obviously also a product of maximal ideals, and we have I2 ⊂ I. Hence I2 = I bythe minimality of I.

(ii) If PER is any maximal ideal, then PI is also a product of maximal ideals with PI ⊂ I.So again by the minimality of I we see that I = PI, which is clearly contained in P.Hence I is contained in every maximal ideal of R.

Now let us assume for a contradiction that I 6= 0. Then, as R is Artinian, Lemma 7.4 (b)implies that there is a minimal ideal JER with IJ 6= 0. About this ideal we note:

• J is a principal ideal (so in particular finitely generated): there must be an elementb ∈ J with I · (b) 6= 0, and we clearly have (b)⊂ J, so (b) = J by the minimality of J.

• IJ = J again by the minimality of J, since IJ ⊂ J and I · IJ = I2J = IJ 6= 0 by (i).

Because of these two properties of J Nakayama’s Lemma in Corollary 3.27 gives us anelement a ∈ I with (1−a)J = 0. As J 6= 0, this means that 1−a is not a unit in R. But then(1−a) 6= R, hence 1−a is contained in a maximal ideal PER. But so is a ∈ I ⊂ P by (ii),and thus we obtain the contradiction 1 = (1−a)+a ∈ P. Hence we conclude that I = 0.

(b) Let PER be any prime ideal. Then Pi ⊂ P for some i = 1, . . . ,n, since otherwise there wouldbe elements ai ∈ Pi\P for all i, which implies a1 · · · · ·an ∈ P1 · · · · ·Pn = 0⊂ P although noai lies in P. But Pi is maximal, and so we must have P = Pi. �

68 Andreas Gathmann

A somewhat surprising consequence of this proposition is that every Artinian ring is Noetherian —a statement that is false for general modules as we have seen in Exercise 7.3:

Proposition 7.17 (Hopkins). For any ring R we have:

R is Artinian ⇔ R is Noetherian and every prime ideal of R is maximal.

Proof.

“⇒” Let R be Artinian. Then every prime ideal is maximal by Proposition 7.16 (b). Moreover, byProposition 7.16 (a) there are maximal ideals P1, . . . ,Pn of R giving a chain

0 = Q0 ⊂ Q1 ⊂ ·· · ⊂ Qn = R

of ideals in R, where Qi = Pi+1 ·Pi+2 · · · · ·Pn for i = 0, . . . ,n. Now for all i = 1, . . . ,n thequotient Qi/Qi−1 = Qi/PiQi is an Artinian R-module by Lemma 7.7 (b), hence an ArtinianR/Pi-vector space by Example 7.5 (c), therefore also a Noetherian R/Pi-vector space byExample 7.2 (a), and thus a Noetherian R-module again by Example 7.5 (c). So by inductionon i it follows from Lemma 7.7 (a) that Qi is a Noetherian R-module for all i = 0, . . . ,n. Inparticular, R = Qn is Noetherian.

“⇐” Assume that R is Noetherian, but not Artinian. We have to find a prime ideal P of R that isnot maximal.

Consider the family of all ideals I of R such that R/I is not Artinian (as a ring or as anR-module, see Example 7.5 (c)). This family is non-empty since it contains the zero ideal,so as R is Noetherian it must have a maximal element P by Lemma 7.4 (a). Note that P iscertainly not a maximal ideal: otherwise R/P would be a field by Lemma 2.3 (b), henceArtinian by Example 7.2 (a) — in contradiction to the choice of P.

It therefore suffices to prove that P is prime, i. e. by Lemma 2.3 (a) that S := R/P is anintegral domain. For any a ∈ R consider the exact sequence

0−→ S/ann(a) ·a−→ S−→ S/(a)−→ 0

of S-modules. As S is not Artinian, we know by Remark 7.8 (a) that at least one of the ringsS/ann(a) and S/(a) cannot be Artinian either. But since P was chosen to be maximal suchthat S = R/P is not Artinian, taking a further quotient of S by a non-zero ideal must yield anArtinian ring, and thus we conclude that ann(a) = 0 or a = 0. In other words, every non-zeroelement of R/P is a non-zero-divisor, i. e. R/P is an integral domain. �

Example 7.18. Let I be a non-zero ideal in a principal ideal domain R. Then as in Example 1.4we have I = (a) with a = pa1

1 · · · · · pann , where a1, . . . ,an ∈ N>0 and p1, . . . , pn are distinct prime

elements of R. We want to check the conditions of Proposition 7.17 for the ring S := R/I.

First of all, by Lemma 1.21 the ideals of S correspond to the ideals of R that contain I. These are theideals (b) with b |a, i. e. such that b = pb1

1 · · · · · pbnn with bi ≤ ai for all i. In particular, S has only

finitely many ideals, which means by Definition 7.1 that S is trivially Noetherian as well as Artinian.

Moreover, since R is a principal ideal domain we have already seen in Example 2.6 (b) that everynon-zero prime ideal in R is maximal, and hence by Corollary 2.4 that every prime ideal in S ismaximal as well. Hence all three conditions of Proposition 7.17 are satisfied for S. In fact, themaximal ideals of S are just the ideals (pi) for i = 1, . . . ,n. So the equation (p1)

a1 · · · · · (pn)an = 0

also verifies the statement of Proposition 7.16 (a).

Example 7.19. Specializing Example 7.18 to the geometric case R = C[x], we see that

S = C[x]/( f ) with f = (x− x1)a1 · · · · · (x− xn)

an

is an Artinian ring, where x1, . . . ,xn ∈ C are distinct and a1, . . . ,an ∈ N>0. In fact, X = V ( f ) ={x1, . . . ,xn}⊂C is a finite collection of points, in accordance with Remark 7.15. But S is not reducedunless a1 = · · · = an = 1 since the class of (x− x1) · · ·(x− xn) is nilpotent in S, and consequently Sis not the coordinate ring of X . Instead, the ring S remembers the local multiplicity information ateach point, i. e. the exponents a1, . . . ,an in f .


So even if S corresponds to a finite collection of points, the structure of S is not completely deter-mined by giving these points: it also contains some local information at each of these points. Thecorresponding precise algebraic statement is that an Artinian ring is completely determined by itslocalizations at all maximal ideals:

Proposition 7.20 (Structure Theorem for Artinian rings). Every Artinian ring R is a finite productof local Artinian rings.

More precisely, if P1, . . . ,Pn are the distinct maximal ideals of R (see Proposition 7.16 (b)), then thelocalizations RPi are also Artinian for all i = 1, . . . ,n, and the ring R is isomorphic to RP1×·· ·×RPn .

Proof. By Proposition 7.16 we can find k ∈ N such that Pk1 · · · · ·Pk

n = 0. Note that Pk1 , . . . ,P

kn are

pairwise coprime by Exercise 2.24. Hence Pk1 ∩ ·· · ∩Pk

n = Pk1 · · · · ·Pk

n = 0 by Exercise 1.8, and sothe Chinese Remainder Theorem of Proposition 1.14 implies that

R ∼= R/Pk1 ×·· ·×R/Pk

n .

As the factors R/Pki are clearly Artinian by Lemma 7.7 (b), it therefore only remains to be shown that

the ring R/Pki is isomorphic to the localization RPi for all i. In fact, it is straightforward to construct

mutually inverse ring homomorphisms, without loss of generality for i = 1:

• The ring homomorphism R→ RP1 , a 7→ a1 contains Pk

1 in its kernel: if a ∈ Pk1 we can choose

a j ∈ Pj\P1 for all j = 2, . . . ,n. Then u := ak2 · · · · ·ak

n /∈ P1 since P1 is prime, and ua = a ·ak2 ·

· · · ·akn ∈ Pk

1 · · · · ·Pkn = 0. This means that a

1 = 0 in RP1 . Hence the above map gives rise toa ring homomorphism R/Pk

1 → RP1 , a 7→ a1 .

• Now consider the ring homomorphism R→ R/Pk1 , a 7→ a. It maps any a ∈ R\P1 to a unit:

otherwise (a) would be contained in a maximal ideal of R/Pk1 , which must be of the form

P/Pk1 for a maximal ideal P ⊃ Pk

1 of R by Lemma 1.21 and Corollary 2.4. As P is primethis means that P ⊃ P1, and hence P = P1 since P1 is maximal. But a ∈ P1/Pk

1 impliesa ∈ P1, a contradiction. By Exercise 6.12 we conclude that the above map gives us a ringhomomorphism RP1 → R/Pk

1 with a1 7→ a. �

Example 7.21. Let us continue Example 7.18, i. e. consider the ring S = R/I for a principal idealdomain R and I = (a) with a = pa1

1 · · · · · pann . In the proof of Proposition 7.20 we can then take any

k ∈ N with k ≥ ai for all i = 1, . . . ,n, and obtain

S/(pi)k ∼= R/(a, pk

i ) = R/(paii )

by Example 1.4 (a). So the decomposition of S into local Artinian rings given by Proposition 7.20 isjust

S ∼= R/(pa11 )×·· ·×R/(pan

n ),

which by the proposition is also isomorphic to S(p1)×·· ·×S(pn).

Exercise 7.22. Let R be a Noetherian ring. Show:

(a) If R is an integral domain, every non-zero non-unit a ∈ R can be written as a product ofirreducible elements of R.

(b) For any ideal IER there is an n ∈ N such that (√

I)n ⊂ I.

Exercise 7.23. Let S be a multiplicatively closed subset of a ring R. If R is Noetherian (resp.Artinian), show that the localization S−1R is also Noetherian (resp. Artinian).

Exercise 7.24. Prove for any R-module M:

(a) If M is Noetherian then R/annM is Noetherian as well.

(b) If M is finitely generated and Artinian, then M is also Noetherian.

(Hint: You can reduce this to the statement of Proposition 7.17 that an Artinian ring isNoetherian.)

70 Andreas Gathmann

8. Prime Factorization and Primary Decompositions

When it comes to actual computations, Euclidean domains (or more generally principal ideal do-mains) are probably the “nicest” rings that are not fields. One of the main reasons for this is thattheir elements admit a unique prime factorization [G1, Proposition 11.9]. This allows for an easycomputation of many concepts in commutative algebra, such as e. g. the operations on ideals inExample 1.4.

Unfortunately however, such rings are rather rare in practice, with the integers Z and the polynomialring K[x] over a field K being the most prominent examples. So we now want to study in this chapterif there are more general rings that allow a prime factorization of their elements, and what we canuse as a substitute in rings that do not admit such a factorization.

More precisely, given an element a 6= 0 in an integral domain R which is not a unit we ask if we canwrite a = p1 · · · · · pn for some n ∈ N>0 and p1, . . . , pn ∈ R such that:

• The pi for i= 1, . . . ,n are irreducible or prime — recall that in a principal ideal domain thesetwo notions are equivalent, but in a general integral domain we only know that every primeelement is irreducible [G1, Lemma 11.3 and Proposition 11.5].

• The decomposition is unique up to permutation and multiplication with units, i. e. if we alsohave a = q1 · · · · ·qm with q1, . . . ,qm irreducible resp. prime, then m = n and there are unitsc1, . . . ,cn ∈ R and a permutation σ ∈ Sn such that qi = ci pσ(i) for all i = 1, . . . ,n.

Let us first discuss the precise relation between the different variants of these conditions.

Proposition and Definition 8.1 (Unique factorization domains). For an integral domain R the fol-lowing statements are equivalent:

(a) Every non-zero non-unit of R is a product of prime elements.

(b) Every non-zero non-unit of R is a product of irreducible elements, and this decomposition isunique up to permutation and multiplication with units.

(c) Every non-zero non-unit of R is a product of irreducible elements, and every irreducibleelement is prime.

If these conditions hold, R is called factorial or a unique factorization domain (short: UFD).13

Proof.

(a)⇒ (b): Let a ∈ R be a non-zero non-unit. By assumption we know that a = p1 · · · · · pn forsome prime elements p1, . . . , pn. As prime elements are irreducible [G1, Lemma 11.3], wetherefore also have a decomposition into irreducible elements.

Moreover, let a = q1 · · · · ·qm be another decomposition into irreducible elements. Then p1divides a = q1 · · · · ·qm, and as p1 is prime this means that p1 divides one of these factors,without loss of generality p1 |q1. Hence q1 = c p1 for some c ∈ R. But q1 is irreducible andp1 is not a unit, so c must be a unit. This means that p1 and q1 agree up to multiplicationwith a unit. Canceling p1 in the equation p1 · · · · · pn = q1 · · · · · qm by p1 now yieldsp2 · · · · · pn = c ·q2 · · · · ·qm, and continuing with this equation in the same way for p2, . . . , pngives the desired uniqueness statement.

(b)⇒ (c): Let p∈R be irreducible, we have to show that p is prime. So let p |ab, i. e. ab= pc forsome c ∈ R. By assumption we can write all these four elements as products of irreducibleelements and thus obtain an equation

a1 · · · · ·an ·b1 · · · · ·bm = p · c1 · · · · · cr.

8. Prime Factorization and Primary Decompositions 71

But by the uniqueness assumption, the factor p on the right must up to a unit be one of thea1, . . . ,an or b1, . . . ,bm, which implies that p |a or p |b.

(c)⇒ (a) is trivial. �

Remark 8.2. In Proposition 8.1, the assumption in (b) and (c) that every non-zero non-unit can bewritten as a product of irreducible elements is a very weak one: it is satisfied e. g. in every Noetheriandomain by Exercise 7.22 (a). The other conditions are much stronger, as we will see in Examples8.3 (b) and 8.7.

Example 8.3. The following two examples are already known from the “Algebraic Structures” class:

(a) As mentioned above, principal ideal domains (so in particular Z and univariate polynomialrings over a field) are unique factorization domains [G1, Proposition 11.9].

(b) In the ring R = Z[√

5 i] ⊂ C the element 2 obviously divides 6 = (1+√

5 i)(1−√

5 i), butneither 1+

√5 i nor 1−

√5 i. Hence 2 is not prime. But one can show that 2 is irreducible

in R, and thus R is not a unique factorization domain [G1, Example 11.4]. In fact, 2 · 3 =(1+√

5 i)(1−√

5 i) are two decompositions of the same number 6 that do not agree up topermutation and units.

It follows by (a) that R cannot be a principal ideal domain. One can also check this directly:the ideal (2,1+

√5 i) is not principal [G1, Exercise 10.37]. In fact, we will see in Example

13.28 that up to multiplication with a constant this is the only non-principal ideal in R.

We will see in Example 13.8 however that R admits a “unique prime factorization” for ideals(instead of for elements) — a property that holds more generally in so-called Dedekinddomains that we will study in Chapter 13.

Remark 8.4. The most important feature of the unique factorization property is that it is preservedwhen passing from a domain R to the polynomial ring R[x]. Often this is already shown in the“Introduction to Algebra” class, and so we will only sketch the proof of this statement here. It relieson the well-known Lemma of Gauß stating that an irreducible polynomial over Z (or more generallyover a unique factorization domain R) remains irreducible when considered as a polynomial over Q(resp. the quotient field K = QuotR). More precisely, if f ∈ R[x] is reducible in K[x] and factors asf = gh with non-constant g,h ∈ K[x], then there is an element c ∈ K\{0} such that cg and h

c lie inR[x], and so f = (cg) · h

c is already reducible in R[x] [G3, Proposition 3.2 and Remark 3.3].

Proposition 8.5. If R is a unique factorization domain, then so is R[x].

Proof sketch. We will check condition (c) of Definition 8.1 for R[x].

Let f ∈ R[x], and let c be a greatest common divisor of all coefficients of f . Then f = c f ′ forsome polynomial f ′ ∈ R[x] with coefficients whose greatest common divisor is 1, so that no constantpolynomial which is not a unit can divide f ′. Now split f ′ into factors until all of them are irreducible— this process has to stop for degree reasons as we have just seen that the degree of each factor mustbe at least 1. But c can also be written as a product of irreducible elements since R is a uniquefactorization domain, and so f = c f ′ is a product of irreducible elements as well.

Next assume that f is irreducible, in particular we may assume that c = 1. We have to show that fis also prime. So let f divide gh for some g,h ∈ R[x]. If we denote by K the quotient field of R, thenf is also irreducible in K[x] by Remark 8.4, hence prime in K[x] by Example 8.3 (a), and so withoutloss of generality f |g in K[x]. This means that g = f k for some k ∈K[x]. But now by Remark 8.4 wecan find a

b ∈ K (with a and b coprime) such that ab f and b

a k are in R[x]. Since the greatest commondivisor of the coefficients of f is 1 this is only possible if b is a unit. But then k = ab−1 ( b

a k) ∈ R[x],and so f |g in R[x]. �

Remark 8.6. Of course, Proposition 8.5 implies by induction that R[x1, . . . ,xn] = R[x1][x2] · · · [xn] isa unique factorization domain if R is. In particular, the polynomial ring K[x1, . . . ,xn] over a field Kis a unique factorization domain. This also shows that there are more unique factorization domains

72 Andreas Gathmann

than principal ideal domains: as the ideal (x1, . . . ,xn) in K[x1, . . . ,xn] cannot be generated by fewerthan n elements by Exercise 1.9, this polynomial ring is a principal ideal domain only for n = 1.

Example 8.7 (Geometric interpretation of prime and irreducible elements). Consider the coordinatering R = R[x,y]/(x2 + y2− 1) of the unit circle X = V (x2 + y2− 1) ⊂ A2

R. Note that x2 + y2− 1 isobviously irreducible (it cannot be written as a product of two linear polynomials since otherwiseX would have to be a union of two lines), and hence prime by Proposition 8.1 since R[x,y] is aunique factorization domain by Remark 8.6. So (x2 + y2− 1) is a prime ideal by Example 2.6 (a),and consequently R is an integral domain by Lemma 2.3 (a).

We are going to show that R is not a unique factorization domain. In fact, we willprove — and interpret geometrically — that x ∈ R is irreducible, but not prime.Note that the zero locus V (x) of x in X consists of the two points a = (0,1) andb = (0,−1) shown in the picture on the right. In particular, x is neither 0 in R(otherwise V (x) would be X) nor a unit (otherwise V (x) would be empty).

(a) x is not prime: Geometrically, by Remark 2.7 (b) this is just the state-ment that V (x) is not an irreducible variety since it consists of twopoints.

b

a

X

Algebraically, x divides x2 = (1+ y)(1− y) in R, but it does not divide 1± y: if e. g. we hadx |1+ y this would mean 1+ y = gx+h(x2 + y2−1) for some g,h ∈ R[x,y], but plugging inthe point a would then give the contradiction 2 = 0.

(b) x is irreducible: otherwise we would have x = f g for two non-units f and g in R.

Intuitively, as the function x vanishes on X exactly at the two points a and b with multiplicity1, this would mean that one of the two factors, say f , would have to vanish exactly at awith multiplicity 1, and the other g exactly at b. But this would mean that the curve V ( f ) inA2R meets the circle X exactly at one point a with multiplicity 1. This seems geometrically

impossible since the circle X has an outside and an inside, so if V ( f ) crosses the circle andgoes from the outside to the inside, it has to cross it again somewhere (as the dashed line inthe picture above) since it cannot end in the interior of the circle.

To give an exact argument for this requires a bit more work. Note that every element h ∈ Rhas a unique representative of the form h0 + xh1 ∈ R[x,y] with h0,h1 ∈ R[y]. We define a“norm” map

N : R→ R[y], h 7→ h20 +(y2−1)h2

1

which can also be thought of as taking the unique representative of h(x,y) · h(−x,y) notcontaining x. In particular, N is multiplicative (which can of course also be checked directly).Hence we have

(y+1)(y−1) = N(x) = N( f )N(g).

As R[y] is a unique factorization domain, there are now two possibilities (up to symmetry inf and g):

• N( f ) is constant: Then f 20 +(y2− 1) f 2

1 is constant. But the leading coefficients ofboth f 2

0 and (y2−1) f 21 are non-negative and thus cannot cancel in the sum, and hence

we must have that f0 is constant and f1 = 0. But then f is a unit in R, which weexcluded.

• N( f ) = a(y− 1) for some a ∈ R\{0}: Then f 20 +(y2− 1) f 2

1 = a(y− 1), and so wehave y− 1 | f0. So we can write f0 = (y− 1) f ′0 for some polynomial f ′0 ∈ R[y] andobtain (y− 1) f ′0

2 +(y+ 1) f 21 = a. This is again a contradiction, since the left hand

side must have a positive non-constant leading term.

Altogether, this contradiction shows that x is in fact irreducible.

Exercise 8.8. In contrast to Example 8.7, show that the following rings are unique factorizationdomains:


(a) the coordinate rings R[x,y]/(y−x2) and R[x,y]/(xy−1) of the standard parabola and hyper-bola in A2

R, respectively;

(b) C[x,y]/(x2 + y2−1).

We will also see in Proposition 12.14 “(e)⇒ (c)” that the localization of R[x,y]/(x2 + y2−1) at themaximal ideal corresponding to one of the points a and b in Example 8.7 is a unique factorizationdomain. For the moment however we just note that the unique factorization property easily breaksdown, and in addition is only defined for integral domains — so let us now study how the conceptof prime factorization can be generalized to a bigger class of rings.

To see the idea how this can be done, let us consider Example 8.7 again. The function x ∈ R was notprime since its zero locus V (x) was a union of two points. We could not decompose x as a productof two functions vanishing at only one of the points each, but we can certainly decompose the ideal(x) into two maximal (and hence prime) ideals as

(x) = I(a)∩ I(b) = (x,y−1)∩ (x,y+1),

which by Remark 1.12 is just the algebraic version of saying that V (x) is the union of the two pointsa and b. So instead of elements we should rather decompose ideals of R, in the sense that we writethem as intersections of “easier” ideals. (Note that in the above example we could also have takenthe product of the two ideals instead of the intersection, but for general rings intersections turn out tobe better-behaved, in particular under the presence of zero-divisors. Decompositions into productsof maximal or prime ideals will be studied in Chapter 13, see e. g. Proposition 13.7 (b) and Example13.12.)

To see what these “easier” ideals should be, consider the simple case of a principal ideal domainR: any non-zero ideal IER can be written as I = (pk1

1 · · · · · pknn ) for some distinct prime elements

p1, . . . , pn ∈ R and k1, . . . ,kn ∈ N>0 by Example 8.3 (a), and the “best possible” decomposition ofthis ideal as an intersection of easier ideals is

I = (p1)k1 ∩·· ·∩ (pn)

kn .

So it seems that we are looking for decompositions of ideals as intersections of powers of primeideals. Actually, whereas this is the correct notion for principal ideal domains, we need a slightvariant of prime powers for the case of general rings:

Definition 8.9 (Primary ideals). Let R be a ring. An ideal QER with Q 6= R is called primary if forall a,b ∈ R with ab ∈ Q we have a ∈ Q or bn ∈ Q for some n ∈ N (which is obviously equivalent toa ∈ Q or b ∈

√Q).

Example 8.10 (Primary ideals = powers of prime ideals in principal ideal domains). In a principalideal domain R, the primary ideals are in fact exactly the ideals of the form (pn) for a prime elementp ∈ R and n ∈ N>0:

• The ideal (pn) is primary: if ab ∈ (pn) then ab = cpn for some c ∈ R. But now the n factorsof p are either all contained in a (in which case a∈ (pn)), or at least one of them is containedin b (in which case bn ∈ (pn)).

• Conversely, let I = (pk11 · · · · · pkn

n ) be any primary ideal, where p1, . . . , pn are distinct primesand k1, . . . ,kn ∈N>0. Then we must have n = 1, since otherwise pk1

1 · (pk22 · · · · · pkn

n ) ∈ I, butneither pk1

1 nor any power of pk22 · · · · · pkn

n are in I.

Note that if R is only a unique factorization domain the same argument still works for principal ideals— but not for arbitrary ideals as we will see in Example 8.13 (a).

Remark 8.11. Let R be a ring.

(a) Obviously, every prime ideal in R is primary, but the converse does not hold by Example8.10.

74 Andreas Gathmann

(b) However, if QER is primary then P =√

Q is prime: if ab ∈ P then (ab)n ∈ Q for somen ∈ N. Hence an ∈ Q or bn ∈

√Q, which means that a or b lie in

√Q = P. In fact, P is then

the smallest prime ideal containing Q by Lemma 2.21.

If we want to specify the underlying prime ideal P =√

Q of a primary ideal Q we often saythat Q is P-primary.

(c) The condition of an ideal QER with Q 6= R being primary can also be expressed in terms ofthe quotient ring R/Q: obviously, Definition 8.9 translates into the requirement that ab = 0implies a = 0 or b n = 0 for some n ∈ N. This means for every element b that b n = 0 forsome n ∈ N if there is an exists a 6= 0 with ab = 0. So Q is primary if and only if everyzero-divisor of R/Q is nilpotent.

As in Corollary 2.4 this means that primary ideals are preserved under taking quotients, i. e.for an ideal I ⊂ Q we have that Q is primary in R if and only if Q/I is primary in R/I.

To obtain more examples of primary ideals, we need the following lemma. It gives a particularlyeasy criterion to detect a primary ideal Q if its underlying prime ideal

√Q is maximal.

Lemma 8.12 (Primary ideals over maximal ideals). Let P be a maximal ideal in a ring R. If an idealQER satisfies one of the following conditions:

(a)√

Q = P;

(b) Pn ⊂ Q⊂ P for some n ∈ N>0;

then Q is P-primary.

Proof.

(a) The given condition means that in R/Q the nilradical√

(0) is equal to P/Q, hence maximalby Corollary 2.4. Exercise 2.25 then implies that every element of R/Q is either a unit ornilpotent. Therefore every zero-divisor of R/Q (which is never a unit) is nilpotent, and so Qis primary by Remark 8.11 (c).

(b) Taking radicals in the given inclusions yields√

P =√

Pn ⊂√

Q ⊂√

P, and hence we get√Q =√

P = P. So the result then follows from (a). �

Example 8.13 (Primary ideals 6= powers of prime ideals). In general, primary ideals and powers ofprime ideals are different objects:

(a) Primary ideals need not be powers of prime ideals: let Q = (x2,y) and P = (x,y) in R[x,y].Then

√Q = P is maximal by Example 2.6 (c), and so Q is P-primary by Lemma 8.12 (a).

However, (x2,xy,y2) = P2 ( Q ( P = (x,y), hence Q is not a power of P.

(b) Powers of prime ideals need not be primary: Let R = R[x,y,z]/(xy− z2) and P = (x,z)ER.Then P is prime by Lemma 2.3 (a) since R/P ∼= R[y] is an integral domain. But the powerP2 = (x2,xz,z2) is not primary as xy = z2 ∈ P2, but neither is x in P2 nor any power of y.

However, if R is Noetherian and Q primary with√

Q = P, then Q always contains a power of theprime ideal P by Exercise 7.22 (b).

Remark 8.14. Note that the condition of Definition 8.9 for a primary ideal Q is not symmetric in thetwo factors a and b, i. e. it does not say that ab ∈ Q implies that one of the two factors a and b lie in√

Q. In fact, Example 8.13 (b) shows that this latter condition is not equivalent to Q being primaryas it is always satisfied by powers of prime ideals: if P is prime and ab ∈ Pn for some n ∈ N>0 thenwe also have ab ∈ P, hence a ∈ P or b ∈ P, and so a or b lie in P =

√P =√

Pn.

Let us now prove that every ideal in a Noetherian ring can be decomposed as an intersection ofprimary ideals.

Definition 8.15 (Primary decompositions). Let I be an ideal in a ring R. A primary decompositionof I is a finite collection {Q1, . . . ,Qn} of primary ideals such that I = Q1∩·· ·∩Qn.


Proposition 8.16 (Existence of primary decompositions). In a Noetherian ring every ideal has aprimary decomposition.

Proof. Assume for a contradiction that R is a Noetherian ring that has an ideal without primarydecomposition. By Lemma 7.4 (a) there is then an ideal IER which is maximal among all ideals inR without a primary decomposition. In the quotient ring S := R/I the zero ideal I/I is then the onlyone without a primary decomposition, since by Remark 8.11 (c) contraction and extension by thequotient map give a one-to-one correspondence between primary decompositions of an ideal J ⊃ Iin R and primary decompositions of J/I in R/I.

In particular, the zero ideal (0)E S is not primary itself, and so there are a,b ∈ S with ab = 0, buta 6= 0 and bn 6= 0 for all n ∈ N. Now as R is Noetherian, so is S by Remark 7.8 (b), and hence thechain of ideals

ann(b)⊂ ann(b2)⊂ ann(b3)⊂ ·· ·becomes stationary, i. e. there is an n ∈ N such that ann(bn) = ann(bn+1).

Note that (a) 6= 0 and (bn) 6= 0 by our choice of a and b. In particular, these two ideals have aprimary decomposition. Taking the primary ideals of these two decompositions together, we thenobtain a primary decomposition of (a)∩ (bn) as well. But (a)∩ (bn) = 0: if x ∈ (a)∩ (bn) thenx = ca and x = dbn for some c,d ∈ S. As ab = 0 we then have 0 = cab = xb = dbn+1, henced ∈ ann(bn+1) = ann(bn), which means that x = dbn = 0. This is a contradiction, since the zeroideal in S does not have a primary decomposition by assumption. �

14

Example 8.17.

(a) In a unique factorization domain R every principal ideal I = (pk11 · · · · · pkn

n ) has a primarydecomposition

I = (p1)k1 ∩·· ·∩ (pn)

kn

by Example 8.10 (where p1, . . . , pn are distinct prime elements and k1, . . . ,kn ∈ N>0).

(b) Geometrically, if I = Q1∩·· ·∩Qn is a primary decomposition of an ideal I in the coordinatering of a variety X , we have

V (I) =V (Q1)∪·· ·∪V (Qn) =V (P1)∪·· ·∪V (Pn)

by Remark 1.12, where Pi =√

Qi for i = 1, . . . ,n. So by Remark 2.7 we have decomposedthe subvariety Y =V (I) as a union of irreducible subvarieties V (Pi). As coordinate rings ofvarieties are always Noetherian by Remark 7.15, Proposition 8.16 asserts that such a decom-position of a (sub-)variety into finitely many irreducible subvarieties is always possible.

However, the primary decomposition of an ideal IEA(X) contains more information that isnot captured in its zero locus V (I) alone: we do not only get subvarieties whose union isV (I), but also (primary) ideals whose zero loci are these subvarieties. These primary idealscan be thought of as containing additional “multiplicity information”: for example, the zerolocus of the ideal I = ((x−a1)

k1 · · · · · (x−an)kn) in R[x] is the subset {a1, . . . ,an} of R —

but the ideal also associates to each point ai a multiplicity ki, and the primary decomposition

I = ((x−a1)k1)∩·· ·∩ ((x−an)

kn)

as in (a) remembers these multiplicities.

In fact, in higher dimensions the additional information at each subvariety is more compli-cated than just a multiplicity. We will not study this here in detail, however we will see aninstance of this in Example 8.23.

Having proven that primary decompositions always exist in Noetherian rings, we now want to see inthe rest of this chapter to what extent these decompositions are unique. However, with our currentdefinitions it is quite obvious that they are far from being unique, since they can be changed in twosimple ways:

76 Andreas Gathmann

Example 8.18 (Non-uniqueness of primary decompositions).(a) We can always add “superfluous ideals” to a primary decomposition, i. e. primary ideals that

are already contained in the intersection of the others. For example, (x2) and (x)∩ (x2) inR[x] are two primary decompositions of the same ideal (x2) by Example 8.10.

(b) In a given primary decomposition we might have several primary ideals with the same un-derlying radical, such as in

(x2,xy,y2) = (x2,y)∩ (x,y2) (∗)

in R[x,y]. Note that this equation holds since the ideals (x2,y), (x,y2), and (x2,xy,y2) containexactly the polynomials without the monomials 1 and x, 1 and y, and without any constantor linear terms, respectively. Moreover, all three ideals have the radical P = (x,y), and hencethey are all P-primary by Lemma 8.12 (a). So (∗) are two different primary decompositionsof the same ideal, in which none of the ideals is superfluous as in (a).

By a slight refinement of the definitions it is actually easy to remove these two ambiguities fromprimary decompositions. To do this, we need a lemma first.

Lemma 8.19 (Intersections of primary ideals). Let P be a prime ideal in a ring R. If Q1 and Q2 aretwo P-primary ideals in R, then Q1∩Q2 is P-primary as well.

Proof. First of all, by Lemma 1.7 (b) we have√

Q1∩Q2 =√

Q1 ∩√

Q2 = P∩P = P. Now letab ∈Q1∩Q2, i. e. ab ∈Q1 and ab ∈Q2. As Q1 and Q2 are P-primary we know that a ∈Q1 or b ∈ P,as well as a ∈Q2 or b ∈ P. This is the same as a ∈Q1∩Q2 or b ∈ P =

√Q1∩Q2. Hence Q1∩Q2 is

P-primary. �

Definition 8.20 (Minimal primary decompositions). Let {Q1, . . . ,Qn} be a primary decompositionof an ideal I in a ring R, and let Pi =

√Qi for i = 1, . . . ,n. Then the decomposition is called minimal

if

(a) none of the ideals is superfluous in the intersection, i. e.⋂

j 6=i Q j 6⊂ Qi for all i;

(b) Pi 6= Pj for all i, j with i 6= j.

Corollary 8.21 (Existence of minimal primary decompositions). If an ideal in a ring has a primarydecomposition, it also has a minimal one.

In particular, in a Noetherian ring every ideal has a minimal primary decomposition.

Proof. Starting from any primary decomposition, leave out superfluous ideals, and replace idealswith the same radical by their intersection, which is again primary with the same radical by Lemma8.19.

The additional statement follows in combination with Proposition 8.16. �

Exercise 8.22. Find a minimal primary decomposition of . . .

(a) the ideal I = (x2) in the ring R = R[x,y]/(x2 + y2−1) (see Example 8.7);

(b) the ideal I = (6) in the ring R = Z[√

5 i] (see Example 8.3 (b));

(c) the ideal I = (x,y) · (y,z) in the ring R[x,y,z].

As a consequence of Corollary 8.21, one is usually only interested in minimal primary decompo-sitions. However, even then the decompositions will in general not be unique, as the followingexample shows.

Example 8.23 (Non-uniqueness of minimal primary decompositions). Let us consider the idealI = (y) · (x,y) = (xy,y2) in R[x,y]. Geometrically, the zero locus of this ideal is just the horizontalaxis V (I) =V (y), which is already irreducible. However, I is not primary since yx ∈ I, but y /∈ I and


xn /∈ I for all n ∈ N. Hence, I is not its own minimal primary decomposition. However, we claimthat

I = Q1∩Q2 = (y)∩ (x2,xy,y2) and I = Q1∩Q′2 = (y)∩ (x,y2)

are two different minimal primary decompositions of I. In fact, both equations can be checked in thesame way as in Example 8.18 (b) (the ideal I contains exactly the polynomials with no monomial1, y, or xn for n ∈ N). Moreover, Q1 is primary since it is prime, and Q2 and Q′2 are primary byLemma 8.12 (a) since both of them have the same maximal radical P2 = (x,y). Finally, it is clear thatin both decompositions none of the ideals is superfluous, and that the radicals of the two ideals aredifferent — namely P1 = (y) with zero locus X1 =V (P1) = R×{0} and P2 = (x,y) with zero locusX2 =V (P2) = {(0,0)}, respectively.

So geometrically, even our minimal primary decompositions contain a so-called embedded compo-nent, i. e. a subvariety X2 contained in another subvariety X1 of the decomposition, so that it is notvisible in the zero locus of I. The other component X1 is usually called an isolated component. Thecorresponding algebraic statement is that P2 contains another prime ideal P1 occurring as a radicalin the decomposition; we will also say that P2 is an embedded and P1 an isolated prime ideal (seeDefinition 8.25 and Example 8.28).

The intuitive reason why this embedded component occurs is that X2 has ahigher “multiplicity” in I than X1 (in a sense that we do not want to makeprecise here). We can indicate this as in the picture on the right by a “fatpoint” X2 on a “thin line” X1.

X2

X1

In any case, we conclude that even minimal primary decompositions of an ideal are not unique.However, this non-uniqueness is very subtle: we will now show that it can only occur in the pri-mary ideals of embedded components. More precisely, we will prove that in a minimal primarydecomposition:

(a) the underlying prime ideals of all primary ideals are uniquely determined (see Proposition8.27); and

(b) the primary ideals corresponding to all isolated components are uniquely determined (seeProposition 8.34).

So in our example above, P1, P2, and Q1 are uniquely determined, and only the primary ideal corre-sponding to P2 can depend on the decomposition.

To prove the first statement (a), we give an alternative way to reconstruct all underlying prime idealsof a minimal primary decomposition (the so-called associated prime ideals) without knowing thedecomposition at all.

Lemma 8.24. Let Q be a P-primary ideal in a ring R. Then for any a ∈ R we have√Q :a =

{R if a ∈ Q,

P if a /∈ Q.

Proof. If a ∈ Q then clearly Q :a = R, and thus√

Q :a = R as well.

Now let a /∈ Q. Then for any b ∈ Q : a we have ab ∈ Q, and so b ∈ P since Q is P-primary. HenceQ⊂ Q :a⊂ P, and by taking radicals we obtain P⊂

√Q :a⊂ P as desired. �

Definition 8.25 (Associated, isolated, and embedded prime ideals). Let I be an ideal in a ring R.

(a) An associated prime ideal of I is a prime ideal that can be written as√

I :a for some a ∈ R.We denote the set of these associated primes by Ass(I).

(b) The minimal elements of Ass(I) are called isolated prime ideals of I, the other ones em-bedded prime ideals of I.

Remark 8.26. For an ideal I of a ring R, note that not every ideal that can be written as√

I :a forsome a ∈ R is prime. By definition, Ass(I) contains only the prime ideals of this form.

78 Andreas Gathmann

Proposition 8.27 (First Uniqueness Theorem for primary decompositions). Let Q1, . . . ,Qn form aminimal primary decomposition for an ideal I in a ring R, and let Pi =

√Qi for i = 1, . . . ,n.

Then {P1, . . . ,Pn} = Ass(I). In particular, the number of components in a minimal primary decom-position and their radicals do not depend on the chosen decomposition.

Proof.

“⊂”: Let i ∈ {1, . . . ,n}, we will show that Pi ∈ Ass(I). As the given decomposition is minimal,we can find a ∈

⋂j 6=i Q j with a /∈ Qi. Then√

I :a =√

Q1 :a∩·· ·∩Qn :a

=√

Q1 :a∩·· ·∩√

Qn :a (Lemma 1.7 (b))= Pi (Lemma 8.24),

and so Pi ∈ Ass(I).

“⊃”: Let P ∈ Ass(I), so P =√

I :a for some a ∈ R. Then as above we have

P =√

I :a =√

Q1 :a∩·· ·∩√

Qn :a,

and thus P⊃√

Qi :a for some i by Exercise 2.10 (a) since P is prime. But of course the aboveequation also implies P⊂

√Qi :a, and so P =

√Qi :a. Now by Lemma 8.24 this radical can

only be Pi or R, and since P is prime we conclude that we must have P = Pi. �

Example 8.28. In the situation of Example 8.23, Proposition 8.27 states that the associated primeideals of I are P1 and P2 since we have found a minimal primary decomposition of I with these un-derlying prime ideals. It is then obvious by Definition 8.25 that P1 is an isolated and P2 an embeddedprime of I.

Exercise 8.29. Let I be an ideal in a ring R. In Definition 8.25 we have introduced Ass(I) as the setof prime ideals that are of the form

√I : a for some a ∈ R. If R is Noetherian, prove that we do not

have to take radicals, i. e. that Ass(I) is also equal to the set of all prime ideals that are of the formI : a for some a ∈ R.

Corollary 8.30 (Isolated prime ideals = minimal prime ideals). Let I be an ideal in a Noetherianring R. Then the isolated prime ideals of I are exactly the minimal prime ideals over I as in Exercise2.23, i. e. the prime ideals P⊃ I such that there is no prime ideal Q with I ⊂ Q ( P.

In particular, in a Noetherian ring there are only finitely many minimal prime ideals over any givenideal.

Proof. As R is Noetherian, there is a minimal primary decomposition I = Q1 ∩ ·· · ∩Qn of I byCorollary 8.21. As usual we set Pi =

√Qi for all i, so that Ass(I) = {P1, . . . ,Pn} by Proposition 8.27.

Note that if P ⊃ I is any prime ideal, then P ⊃ Q1 ∩ ·· · ∩Qn, hence P ⊃ Qi for some i by Exercise2.10 (a), and so by taking radicals P ⊃

√Qi = Pi. With this we now show both implications stated

in the corollary:

• Let Pi ∈ Ass(I) be an isolated prime ideal of I. If P is any prime ideal with I ⊂ P ⊂ Pi thenby what we have just said Pj ⊂ P⊂ Pi for some j. But as Pi is isolated we must have equality,and so P = Pi. Hence Pi is minimal over I.

• Now let P be a minimal prime over I. By the above I ⊂ Qi ⊂ Pi ⊂ P for some i. As P isminimal over I this means that P = Pi is an associated prime, and hence also an isolatedprime of I. �

Remark 8.31. In particular, Corollary 8.30 states that the isolated prime ideals of an ideal I in a co-ordinate ring of a variety A(X) correspond exactly to the maximal subvarieties, i. e. to the irreduciblecomponents of V (I) — as already motivated in Example 8.23.

Exercise 8.32. Let R be a Noetherian integral domain. Show:


(a) R is a unique factorization domain if and only if every minimal prime ideal over a principalideal is itself principal.

(b) If R is a unique factorization domain then every minimal non-zero prime ideal of R is prin-cipal.

Finally, to prove the second uniqueness statement (b) of Example 8.23 the idea is to use localizationat an isolated prime ideal to remove from I all components that do not belong to this prime ideal.

Lemma 8.33. Let S be a multiplicatively closed subset in a ring R, and let Q be a P-primary idealin R. Then with respect to the ring homomorphism ϕ : R→ S−1R, a 7→ a

1 we have

(Qe)c =

{R if S∩P 6= /0,Q if S∩P = /0.

Proof. If S∩P 6= /0 there is an element s ∈ S with s ∈ P =√

Q, and thus sn ∈ Q for some n ∈ N. So11 = sn

sn ∈ S−1Q = Qe by Example 6.18. Hence Qe = S−1R, and therefore (Qe)c = R.

On the other hand, assume now that S∩P = /0. By Exercise 1.19 (a) it suffices so prove (Qe)c ⊂ Q.If a ∈ (Qe)c we have a

1 ∈ Qe, and so a1 = q

s for some q ∈ Q and s ∈ S by Proposition 6.7 (a). Henceu(q−as) = 0 for some u ∈ S, which implies that a ·us = uq ∈ Q. As Q is P-primary it follows thata ∈ Q or us ∈ P. But us ∈ S, so us /∈ P since S∩P = /0, and we conclude that a ∈ Q. �

Proposition 8.34 (Second Uniqueness Theorem for primary decompositions). Let Q1, . . . ,Qn forma minimal primary decomposition for an ideal I in a ring R, and let Pi =

√Qi for i = 1, . . . ,n.

If i ∈ {1, . . . ,n} such that Pi is minimal over I, then (Ie)c = Qi, where contraction and extension aretaken with respect to the canonical localization map R→ RPi . In particular, in a minimal primarydecomposition the primary components corresponding to minimal prime ideals do not depend on thechosen decomposition.

Proof. Localizing the equation I = Q1∩·· ·∩Qn at S gives S−1I = S−1Q1∩·· ·∩S−1Qn by Exercise6.24 (b), hence Ie = Qe

1∩ ·· ·∩Qen by Example 6.18, and so (Ie)c = (Qe

1)c∩ ·· ·∩ (Qe

n)c by Exercise

1.19 (d).

Now let Pi be minimal over I, and set S = R\Pi. Then S∩Pi = /0, whereas S∩Pj 6= /0 for all j 6= isince Pj 6⊂ Pi. So applying Lemma 8.33 gives (Ie)c = (Qe

i )c = Qi as desired. �

15

80 Andreas Gathmann

9. Integral Ring Extensions

In this chapter we want to discuss a concept in commutative algebra that has its original motivation inalgebra, but turns out to have surprisingly many applications and consequences in geometry as well.To explain its background, recall from the “Introduction to Algebra” class that the most importantobjects in field theory are algebraic and finite field extensions. More precisely, if K ⊂ K′ is aninclusion of fields an element a ∈ K′ is called algebraic over K if there is a non-zero polynomialf ∈K[x] with coefficients in K such that f (a) = 0. The field extension K ⊂K′ is then called algebraicif every element of K′ is algebraic over K [G3, Definition 2.1].

Of course, for an algebraic element a ∈ K′ over K there is then also a monic polynomial relationan + cn−1an−1 + · · ·+ c0 = 0 for some n ∈ N>0 and c0, . . . ,cn−1 ∈ K, since we can just divide f byits leading coefficient. This trivial statement actually has far-reaching consequences: it means thatwe can use the equation an =−cn−1an−1−·· ·−c0 to reduce any monomial ak for k ∈N (and in factalso the multiplicative inverses of non-zero polynomial expressions in a) to a K-linear combinationof the first n powers 1,a, . . . ,an−1, and that consequently the extension field K(a) of K generatedby a is a finite-dimensional vector space over K — we say that K ⊂ K(a) is a finite field extension[G3, Definition 2.12 and Proposition 2.14 (b)]. This means that the field extension K ⊂K(a) is quiteeasy to deal with, since we can use the whole machinery of (finite-dimensional) linear algebra for itsstudy.

What happens now if instead of an extension K ⊂ K′ of fields we consider an extension R ⊂ R′

of rings? We can certainly still have a look at elements a ∈ R′ satisfying a polynomial relationcnan + cn−1an−1 + · · ·+ c0 = 0 with c0, . . . ,cn ∈ R (and not all of them being 0). But now it will ingeneral not be possible to divide this equation by its leading coefficient cn to obtain a monic relation.Consequently, we can in general not use this relation to express higher powers of a in terms of lowerones, and hence the R-algebra R[a] generated by a over R need not be a finite R-module. A simpleexample for this can already be found in the ring extension Z⊂Q: for example, the rational numbera= 1

2 satisfies a (non-monic) polynomial relation 2a−1= 0 with coefficients in Z, but certainly Z[a],which is the ring of all rational numbers with a finite binary expansion, is not a finitely generatedZ-module.

We learn from this that in the case of a ring extension R⊂ R′ we should often not consider elementsof R′ satisfying polynomial relations with coefficients in R, but rather require monic relations in thefirst place. So let us start by giving the corresponding definitions.

Definition 9.1 (Integral and finite ring extensions).(a) If R ⊂ R′ are rings, we call R′ an extension ring of R. We will also say in this case that

R⊂ R′ is a ring extension.

Note: sometimes in the literature a ring extension is meant to be any ring homomorphismR→ R′, even if it is not injective (so that R′ is an arbitrary R-algebra as in Definition 1.23(a)).

(b) Let R be a ring. An element a of an extension ring R′ of R is called integral over R if thereis a monic polynomial f ∈ R[x] with f (a) = 0, i. e. if there are n ∈ N>0 and c0, . . . ,cn−1 ∈ Rwith an +cn−1an−1 + · · ·+c0 = 0. We say that R′ is integral over R if every element of R′ isintegral over R.

(c) A ring extension R⊂ R′ is called finite if R′ is finitely generated as an R-module.

Remark 9.2.(a) Note that the usage of the word “integral” for the concept of Definition 9.1 (b) is completely

unrelated to the appearance of the same word in the term “integral domain”.

9. Integral Ring Extensions 81

(b) If R and R′ are fields, the notions of integral and finite ring extensions actually coincidewith those of algebraic and finite field extensions [G3, Definitions 2.1 and 2.12]. In fact,for this case you might already know some of the next few results in this chapter from the“Introduction to Algebra” class, in particular Proposition 9.5 and Lemma 9.6.

Example 9.3. Let R be a unique factorization domain, and let R′ = QuotR be its quotient field. Weclaim that a ∈ R′ is integral over R if and only if a ∈ R.

In fact, it is obvious that any element of R is integral over R, so let us prove the converse. Assumethat a = p

q is integral over R with p and q coprime, i. e. there is a polynomial equation( pq

)n+ cn−1

( pq

)n−1+ · · ·+ c0 = 0

with c0, . . . ,cn−1 ∈ R. We multiply with qn to get

pn + cn−1 pn−1q+ · · ·+ c0qn = 0, i. e. pn =−q(cn−1 pn−1 + · · ·+ c0qn−1).

in R. So q | pn, which is only possible if q is a unit since p and q are coprime. Hence a = pq ∈ R.

So in particular, if R is not a field then R′ 6= R, and hence the ring extension R⊂ R′ is not integral.

Example 9.4 (Geometric examples of integral extensions). Let R = C[x] and R′ = R[y]/( f ) =C[x,y]/( f ), where f ∈ R[y] is a (non-constant) polynomial relation for the additional variable y.Geometrically, we then have R = A(X) and R′ = A(X ′) for X =A1

C and the curve X ′ in A2C given by

the equation f (x,y) = 0. The ring extension map R→ R′ corresponds to the morphism of varietiesπ : X ′ → X , (x,y) 7→ x in the sense of Construction 0.11. In the following figure we show threeexamples of this setting, where we only draw the real points to keep the pictures inside A2

R.

(a) (b) (c)

X ′

X

X ′

X

X ′

X

π π π

f = y2− x2 f = xy−1 f = xy

The subtle but important difference between these examples is that in case (a) the given relationf ∈ R[y] is monic in y, whereas in (b) and (c) it is not (with the leading term being xy). This hasgeometric consequences for the inverse images π−1(x) of points x ∈ X , the so-called fibers of π:

(a) In this case, the generator y of R′ over R is integral since it satisfies the monic relationy2−x2 = 0. In fact, Proposition 9.5 will show that this implies that the whole ring extensionR ⊂ R′ is integral. Geometrically, the monic relation means that plugging in an arbitraryvalue for x will always give a quadratic equation y2− x2 = 0 for y, leading to two points inany fiber π−1(x) (counted with multiplicities).

(b) In this example, y ∈ R′ does not satisfy a monic relation over R: considering the leadingterm in y it is obvious that there are no polynomials g,h ∈ R[y] with g monic in y such thatg = h(xy−1). Hence the extension R ⊂ R′ is not integral. Geometrically, the consequenceis now that after plugging in a value for x the relation xy−1 = 0 for y is linear for x 6= 0 butconstant for x = 0, leading to an empty fiber π−1(0) = /0 whereas all other fibers contain asingle point.

(c) This case is similar to (c): again, the ring extension R ⊂ R′ is not integral, and the relationxy = 0 does not remain linear in y when setting x = 0. This time however, this leads to aninfinite fiber π−1(0) instead of an empty one.

82 Andreas Gathmann

Summarizing, we would expect that a morphism of varieties corresponding to an integral ring exten-sion is surjective with finite fibers. In fact, we will see this in Example 9.19, and thinking of integralextensions geometrically as morphisms with this property is a good first approximation — even ifthe precise geometric correspondence is somewhat more complicated (see e. g. Example 9.25).

But let us now start our rigorous study of integral and finite extensions by proving their main alge-braic properties.

Proposition 9.5 (Integral and finite ring extensions). An extension ring R′ is finite over R if and onlyif R′ = R[a1, . . . ,an] for integral elements a1, . . . ,an ∈ R′ over R.

Moreover, in this case the whole ring extension R⊂ R′ is integral.

Proof.

“⇒”: Let R′ = 〈a1, . . . ,an 〉 be finitely generated as an R-module. Of course, we then also haveR′ = R[a1, . . . ,an], i. e. R′ is generated by the same elements as an R-algebra. We will provethat every element of R′ is integral over R, which then also shows the “moreover” statement.

So let a ∈ R′. As R′ is finite over R, we can apply the Cayley-Hamilton theorem of Proposi-tion 3.25 to the R-module homomorphism ϕ : R′→ R′, x 7→ ax to obtain a monic polynomialequation

ϕk + ck−1ϕ

k−1 + · · ·+ c0 = 0in HomR(R′,R′) for some c0, . . . ,ck−1 ∈ R, and hence ak + ck−1ak−1 + · · ·+ c0 = 0 by plug-ging in the value 1. Thus a is integral over R.

“⇐”: Let R′ = R[a1, . . . ,an] with a1, . . . ,an integral over R, i. e. every ai satisfies a monic polyno-mial relation of some degree ri over R. Now by Lemma 1.28 every element of R′ is a poly-nomial expression of the form ∑k1,...,kn ck1,...,kn ak1

1 · · · · · aknn for some ck1,...,kn ∈ R, and we

can use the above polynomial relations to reduce the exponents to ki < ri for all i = 1, . . . ,n.Hence R′ is finitely generated over R by all monomial expressions ak1

1 · · · · ·aknn with ki < ri

for all i. �

Lemma 9.6 (Transitivity of integral and finite extensions). Let R⊂ R′ ⊂ R′′ be rings.

(a) If R⊂ R′ and R′ ⊂ R′′ are finite, then so is R⊂ R′′.

(b) If R⊂ R′ and R′ ⊂ R′′ are integral, then so is R⊂ R′′.

Proof.

(a) Let a1, . . . ,an generate R′ as an R-module, and b1, . . . ,bm generate R′′ as an R′-module. Thenevery element of R′′ is of the form ∑

mi=1 cibi for some ci ∈ R′, i. e. ∑

mi=1(

∑nj=1 ci, ja j

)·bi for

some ci, j ∈ R. Hence the finitely many products a j bi generate R′′ as an R-module.

(b) Let a ∈ R′′. As a is integral over R′, there are n ∈ N>0 and elements c0, . . . ,cn−1 of R′

such that an + cn−1an−1 + · · ·+ c0 = 0. Then a is also integral over R[c0, . . . ,cn−1]. In ad-dition, we know that c0, . . . ,cn−1 are integral over R. Hence Proposition 9.5 tells us thatR[c0, . . . ,cn−1,a] is finite over R[c0, . . . ,cn−1] and R[c0, . . . ,cn−1] is finite over R. ThereforeR[c0, . . . ,cn−1,a] is finite over R by (a), and thus a is integral over R by Proposition 9.5again. �

A nice property of integral extensions is that they are compatible with quotients, localizations, andpolynomial rings in the following sense.

Lemma 9.7. Let R′ be an integral extension ring of R.

(a) If I is an ideal of R′ then R′/I is an integral extension ring of R/(I∩R).

(b) If S is a multiplicatively closed subset of R then S−1R′ is an integral extension ring of S−1R.

(c) R′[x] is an integral extension ring of R[x].


Proof.

(a) Note that the map R/(I ∩ R)→ R′/I, a 7→ a is well-defined and injective, hence we canregard R′/I as an extension ring of R/(I ∩R). Moreover, for all a ∈ R′ there is a monicrelation an + cn−1an−1 + · · ·+ c0 = 0 with c0, . . . ,cn−1 ∈ R, and hence by passing to thequotient also an + cn−1 an−1 + · · ·+ c0 = 0. So a is integral over R/(I∩R).

(b) Again, the ring homomorphism S−1R→ S−1R′, as 7→

as is obviously well-defined and injec-

tive. Moreover, for as ∈ S−1R′ we have a monic relation an + cn−1an−1 + · · ·+ c0 = 0 with

c0, . . . ,cn−1 ∈ R, and thus also(as

)n+

cn−1

s

(as

)n−1+ · · ·+ c0

sn = 0.

Hence as is integral over S−1R.

(c) Let f = anxn + · · ·+ a0 ∈ R′[x], i. e. a0, . . . ,an ∈ R′. Then a0, . . . ,an are integral over R, soalso over R[x], and thus R[x][a0, . . . ,an] = R[a0, . . . ,an][x] is integral over R[x] by Proposition9.5. In particular, this means that f is integral over R[x]. �

Exercise 9.8.(a) Is

√2+√

2+ 12

3√

3 ∈ R integral over Z?

(b) Let R′ = R[x], R = R[x2− 1] ⊂ R′, P′ = (x− 1)ER′, and P = P′ ∩R. Show that R′ is anintegral extension of R, but the localization R′P′ is not an integral extension of RP. Is this acontradiction to Lemma 9.7 (b)?

(Hint: consider the element 1x+1 .)

An important consequence of our results obtained so far is that the integral elements of a ring exten-sion always form a ring themselves. This leads to the notion of integral closure.

Corollary and Definition 9.9 (Integral closures).(a) Let R⊂ R′ be a ring extension. The set R of all integral elements in R′ over R is a ring with

R⊂ R⊂ R′. It is called the integral closure of R in R′. We say that R is integrally closed inR′ if its integral closure in R′ is R.

(b) An integral domain R is called integrally closed or normal if it is integrally closed in itsquotient field QuotR.

Proof. It is clear that R⊂ R⊂ R′, so we only have to show that R is a subring of R′. But this followsfrom Proposition 9.5: if a,b ∈ R′ are integral over R then so is R[a,b], and hence in particular a+band a ·b. �

Example 9.10. Every unique factorization domain R is normal, since by Example 9.3 the onlyelements of QuotR that are integral over R are the ones in R.

In contrast to Remark 9.2 (b), note that for a field R Definition 9.9 (b) of an integrally closed domaindoes not specialize to that of an algebraically closed field: we do not require that R admits nointegral extensions at all, but only that it has no integral extensions within its quotient field QuotR.The approximate geometric meaning of this concept can be seen in the following example.

Example 9.11 (Geometric interpretation of normal domains). Let R=A(X) be the coordinate ring ofan irreducible variety X . The elements ϕ = f

g ∈ QuotR of the quotient field can then be interpretedas rational functions on X , i. e. as quotients of polynomial functions that are well-defined exceptat some isolated points of X (where the denominator g vanishes). Hence the condition of R beingnormal means that every rational function ϕ satisfying a monic relation ϕn+cn−1ϕn−1+ · · ·+c0 = 0with c0, . . . ,cn−1 ∈ R is already an element of R, so that its value is well-defined at every point of X .Now let us consider the following two concrete examples:

84 Andreas Gathmann

(a) Let R = C[x], corresponding to the variety X = A1C. By Example 9.10 we know that R is

normal. In fact, this can also be understood geometrically: the only way a rational functionϕ on A1

C can be ill-defined at a point a ∈ A1C is that it has a pole, i. e. that it is of the

form x 7→ f(x−a)k for some k ∈ N>0 and f ∈ QuotR that is well-defined and non-zero at a.

But then ϕ cannot satisfy a monic relation of the form ϕn + cn−1ϕn−1 + · · ·+ c0 = 0 withc0, . . . ,cn−1 ∈ C[x] since ϕn has a pole of order kn at a which cannot be canceled by thelower order poles of the other terms cn−1ϕn−1 + · · ·+ c0 = 0. Hence any rational functionsatisfying such a monic relation is already a polynomial function, which means that R isnormal.

(b) Let X =V (y2−x2−x3)⊂A2R and R = A(X) =R[x,y]/(y2−x2−x3).

The curve X is shown in the picture on the right: locally around theorigin (i. e. for small x and y) the term x3 is small compared to x2 andy2, and thus X is approximatively given by (y−x)(y+x)= y2−x2≈ 0,which means that it consists of two branches crossing the origin withslopes ±1.

X

y

x

In this case the ring R is not normal: the rational function ϕ = yx ∈ Quot(R)\R satisfies the

monic equation ϕ2− x−1 = y2

x2 − x−1 = x3+x2

x2 − x−1 = 0. Geometrically, the reason whyϕ is ill-defined at 0 is not that it has a pole (i. e. tends to ∞ there), but that it approachestwo different values 1 and −1 on the two local branches of the curve. This means that ϕ2

approaches a unique value 1 at the origin, and thus has a well-defined value at this point —leading to the monic quadratic equation for ϕ . So the reason for R not being normal is the“singular point” of X at the origin. In fact, one can think of the normality condition geo-metrically as some sort of “non-singularity” statement (see also Example 11.37 and Remark12.15 (b)).

Exercise 9.12 (Integral closures can remove singularities). As in Example 9.11 (b) let us againconsider the ring R =R[x,y]/(y2−x2−x3), and let K = QuotR be its quotient field. For the elementt := y

x ∈ K, show that the integral closure R of R in K is R[t], and that this is equal to R[t]. What isthe geometric morphism of varieties corresponding to the ring extension R⊂ R?

Exercise 9.13. Let R be an integral domain, and let S⊂ R be a multiplicatively closed subset. Prove:

(a) Let R′ ⊃ R be an extension ring of R. If R is the integral closure of R in R′, then S−1R is theintegral closure of S−1R in S−1R′.

(b) If R is normal, then S−1R is normal.

(c) (Normality is a local property) If RP is normal for all maximal ideals PER, then R is normal.

Exercise 9.14. Let R ⊂ R′ be an extension of integral domains, and let R be the integral closure ofR in R′.

Show that for any two monic polynomials f ,g ∈ R′[t] with f g ∈ R[t] we have f ,g ∈ R[t].

(Hint: From the “Introduction to Algebra” class you may use the fact that any polynomial over afield K has a splitting field, so in particular an extension field L⊃ K over which it splits as a productof linear factors [G3, Proposition 4.15].)

16Checking whether an element a∈R′ of a ring extension R⊂R′ is integral over R can be difficult sincewe have to show that there is no monic polynomial over R at all that vanishes at a. The followinglemma simplifies this task if R is a normal domain: it states that in this case it suffices to consideronly the minimal polynomial of a over the quotient field K = QuotR (i. e. the uniquely determinedmonic polynomial f ∈ K[x] of smallest degree having a as a zero [G3, Definition 2.4]), since thispolynomial must already have coefficients in R if a is integral.

Lemma 9.15. Let R⊂ R′ be an integral extension of integral domains, and assume that R is normal.

(a) For any a ∈ R′ its minimal polynomial f over K = QuotR actually has coefficients in R.


(b) If moreover a ∈ PR′ for some prime ideal PER, then the non-leading coefficients of f areeven contained in P.

Proof.

(a) As a is integral over R there is a monic polynomial g ∈ R[x] with g(a) = 0. Then f |g overK [G3, Remark 2.5], i. e. we have g = f h for some h ∈ K[x]. Applying Exercise 9.14 to theextension R⊂ K (and R = R since R is normal) it follows that f ∈ R[x] as required.

(b) Let a = p1a1 + · · ·+ pkak for some p1, . . . , pk ∈ P and a1, . . . ,ak ∈ R′. Replacing R′ byR[a1, . . . ,ak] we may assume by Proposition 9.5 that R′ is finite over R. Then we can applythe Cayley-Hamilton theorem of Proposition 3.25 to ϕ : R′→ R′, x 7→ ax: since the imageof this R-module homomorphism lies in PR′, we obtain a polynomial relation

ϕn + cn−1ϕ

n−1 + · · ·+ c0 = 0 ∈ HomR(R′,R′)

with c0, . . . ,cn−1 ∈ P. Plugging in the value 1 this means that we have a monic polynomialg ∈ R[x] with non-leading coefficients in P such that g(a) = 0.

By the proof of (a) we can now write g = f h, where f ∈ R[x] is the minimal polynomial of aand h ∈ R[x]. Reducing this equation modulo P gives x n = f h in (R/P)[x]. But as R/P is anintegral domain by Lemma 2.3 (a) this is only possible if f and h are powers of x themselves(otherwise the highest and lowest monomial in their product would have to differ). Hencethe non-leading coefficients of f lie in P. �

Example 9.16 (Integral elements in quadratic number fields). Let d ∈ Z\{0,1} be a square-freeinteger. We want to compute the elements in Q(

√d) = {a+ b

√d : a,b ∈ Q} ⊂ C that are integral

over Z, i. e. the integral closure R of R = Z in R′ =Q(√

d). These subrings of C play an importantrole in number theory; you have probably seen them already in the “Elementary Number Theory”class [M, Chapter 8].

It is obvious that the minimal polynomial of a+b√

d ∈Q(√

d)\Q over Q is

(x−a−b√

d)(x−a+b√

d) = x2−2ax+a2−db2.

So as Z is normal by Example 9.10 we know by Lemma 9.15 (a) (applied to the integral extensionR⊂ R) that a+b

√d is integral over Z if and only if this polynomial has integer coefficients. Hence

R = {a+b√

d : a,b ∈Q,−2a ∈ Z,a2−db2 ∈ Z},which is the usual way how this ring is defined in the “Elementary Number Theory” class [M,Definition 8.12]. Note that this is in general not the same as the ring Z+Z

√d — in fact, one can

show that R = Z+Z√

d only if d 6= 1 mod 4, whereas for d = 1 mod 4 we have R = Z+Z 1+√

d2

[M, Proposition 8.16].

Remark 9.17 (Geometric properties of integral ring extensions). After having discussed the alge-braic properties of integral extensions, let us now turn to the geometric ones (some of which werealready motivated in Example 9.4). So although we will continue to allow general (integral) ringextensions R ⊂ R′, our main examples will now be coordinate rings R = A(X) and R′ = A(X ′) ofvarieties X and X ′, respectively. The inclusion map i : R→ R′ then corresponds to a morphism ofvarieties π : X ′→ X as in Construction 0.11 and Example 9.4.

We are going to study the contraction and extension of prime ideals by the ring homomorphism i— by Remarks 1.18 and 2.7 this means that we consider images and inverse images of irreduciblesubvarieties under π . As i is injective, i. e. R is a subring of R′, note that

• the contraction of a prime ideal P′ER′ is exactly (P′)c = P′∩R; and

• the extension of a prime ideal PER is exactly Pe = PR′.

Moreover, by Example 2.9 (b) we know that the contraction P′∩R of a prime ideal P′ER′ is alwaysprime again, which means geometrically that the image of an irreducible subvariety X ′ under π isirreducible. The main geometric questions that we can ask are whether conversely any prime ideal

86 Andreas Gathmann

PER is of the form P′∩R for some prime ideal P′ER′ (maybe with some additional requirementsabout P′), corresponding to studying whether there are irreducible subvarieties of X ′ with givenimage under π (and what we can say about them).

For the rest of this chapter, our pictures to illustrate such questions will alwaysbe of the form as shown on the right: we will draw the geometric varieties andsubvarieties, but label them with their algebraic counterparts. The arrow fromthe top to the bottom represents the map π , or algebraically the contraction mapP′ 7→ P′∩R on prime ideals (with the actual ring homomorphism i going in theopposite direction). For example, in the picture on the right the point V (P′) inX ′ for a maximal ideal P′ER′ is mapped to the point V (P) in X for a maximalideal PER, which means that P′∩R = P.

P′R′

∩R

PR

There are four main geometric results on integral ring extensions in the above spirit; they are com-monly named Lying Over, Incomparability, Going Up, and Going Down. We will treat them now inturn. The first one, Lying Over, is the simplest of them — it just asserts that for an integral extensionR⊂ R′ every prime ideal in R is the contraction of a prime ideal in R′. In our pictures, we will alwaysuse gray color to indicate objects whose existence we are about to prove.

Proposition 9.18 (Lying Over). Let R⊂ R′ be a ring extension, and PER prime.

(a) There is a prime ideal P′ER′ with P′∩R = P if and only if PR′∩R⊂ P.

(b) If R⊂ R′ is integral, then this is always the case.

We say in this case that P′ is lying over P.

P′R′:

R: P

Proof.

(a) “⇒” If P′∩R = P then PR′∩R = (P′∩R)R′∩R⊂ P′R′∩R = P′∩R = P.

“⇐” Consider the multiplicatively closed set S = R\P. As PR′ ∩ S = (PR′ ∩R)\P = /0 byassumption, Exercise 6.14 (a) implies that there is a prime ideal P′ER′ with PR′ ⊂ P′

and P′ ∩ S = /0. But the former inclusion implies P ⊂ PR′ ∩R ⊂ P′ ∩R and the latterP′∩R = P′∩P⊂ P, so we get P′∩R = P as desired.

(b) Let a ∈ PR′ ∩R. Since a ∈ PR′ it follows from the Cayley-Hamilton theorem as in the firsthalf of the proof of Lemma 9.15 (b) that there is a monic relation an+cn−1an−1+ · · ·+c0 = 0with c0, . . . ,cn−1 ∈ P. As moreover a ∈ R, this means that an = −cn−1an−1−·· ·− c0 ∈ P,and thus a ∈ P as P is prime. �

Example 9.19. Let us consider the three ring extensions R′ of R = C[x] from Example 9.4 again.

R′ = C[x,y]/(xy)(c)

R′

R P

Q′

P′

R

R′ P′

(a)R′ = C[x,y]/(y2− x2)

P

R′

R

R′ = C[x,y]/(xy−1)(b)

P

Recall that the extension (a) is integral by Proposition 9.5. Correspondingly, the picture above onthe left shows a prime ideal P′ lying over P, i. e. such that P′ ∩ R = P. In contrast, in case (b)there is no prime ideal lying over P, which means by Proposition 9.18 (b) that the ring extensionR ⊂ R′ cannot be integral (as we have already seen in Example 9.4). In short, over an algebraicallyclosed field the restriction of the Lying Over property to maximal ideals PER (i. e. to points of thecorresponding variety) means that morphisms of varieties corresponding to integral ring extensionsare always surjective.


Of course, a prime ideal P′ER′ lying over a given prime ideal PER is in general not unique —there are e. g. two choices for P′ in example (a) above. The case (c) is different however: as the fiberover P is one-dimensional, there are not only many choices for prime ideals lying over P, but alsosuch prime ideals P′ and Q′ with Q′ ( P′ as shown in the picture above. We will prove now thatsuch a situation cannot occur for integral ring extensions, which essentially means that the fibers ofthe corresponding morphisms have to be finite.

Proposition 9.20 (Incomparability). Let R⊂ R′ be an integral ring extension.If P′ and Q′ are distinct prime ideals in R′ with P′ ∩R = Q′ ∩R then P′ 6⊂ Q′

and Q′ 6⊂ P′.

Proof. Let P′∩R = Q′∩R and P′ ⊂ Q′. We will prove that Q′ ⊂ P′ as well, sothat P′ = Q′.

P′ Q′6⊂6⊃R′:

R: P

Assume for a contradiction that there is an element a ∈ Q′\P′. By Lemma 9.7 (a) we know thatR′/P′ is integral over R/(P′∩R), so there is a monic relation

an + cn−1 an−1 + · · ·+ c0 = 0 (∗)

in R′/P′ with c0, . . . ,cn−1 ∈ R. Pick such a relation of minimal degree n. Since a ∈ Q′ this relationimplies c0 ∈Q′/P′, but as c0 ∈R too we conclude that c0 ∈ (Q′∩R)/(P′∩R)= (Q′∩R)/(Q′∩R)= 0.Hence (∗) has no constant term. But since a 6= 0 in the integral domain R′/P′ we can then divide therelation by a to get a monic relation of smaller degree — in contradiction to the choice of n. �

In geometric terms, the following corollary is essentially a restatement of the finite fiber property:it says that in integral ring extensions only maximal ideals can contract to maximal ideals, i. e. thatpoints are the only subvarieties that can map to a single point in the target space.

Corollary 9.21. Let R⊂ R′ be an integral ring extension.

(a) If R and R′ are integral domains then R is a field if and only if R′ is a field.

(b) A prime ideal P′ER′ is maximal if and only if P′∩R is maximal.

Proof.

(a) “⇒” Assume that R is a field, and let P′ER′ be a maximal ideal. Moreover, consider thezero ideal 0ER′, which is prime since R′ is an integral domain. Both ideals contractto a prime ideal in R by Exercise 2.9, hence to 0 since R is a field. Incomparability asin Proposition 9.20 then implies that P′ = 0. So 0 is a maximal ideal of R′, and thus R′

is a field.

“⇐” Now assume that R is not a field. Then there is a non-zero maximal ideal PER. ByLying Over as in Proposition 9.18 (b) there is now a prime ideal P′ER′ with P′∩R=P,in particular with P′ 6= 0. So R′ has a non-zero prime ideal, i. e. R′ is not a field.

(b) By Lemmas 2.3 (a) and 9.7 (a) we know that R/(P′∩R)⊂ R′/P′ is an integral extension ofintegral domains, so the result follows from (a) with Lemma 2.3 (b). �

Exercise 9.22. Which of the following extension rings R′ are integral over R = C[x]?

(a) R′ = C[x,y,z]/(z2− xy);

(b) R′ = C[x,y,z]/(z2− xy,y3− x2);

(c) R′ = C[x,y,z]/(z2− xy,x3− yz).17

Remark 9.23. In practice, one also needs “relative versions” of the LyingOver property: let us assume that we have an (integral) ring extension R⊂ R′

and two prime ideals P⊂ Q in R. If we are now given a prime ideal P′ lyingover P or a prime ideal Q′ lying over Q, can we fill in the other prime ideal sothat P′ ⊂ Q′ and we get a diagram as shown on the right?

P′ Q′R′: ⊂

R: P ⊂ Q

88 Andreas Gathmann

Although these two questions look symmetric, their behavior is in fact rather different. Let us startwith the easier case — the so-called Going Up property — in which we prescribe P′ and are lookingfor the bigger prime ideal Q′. It turns out that this always works for integral extensions. The proofis quite simple: we can reduce it to the standard Lying Over property by first taking the quotient byP′ since this keeps only the prime ideals containing P′ by Lemma 1.21.

Proposition 9.24 (Going Up). Let R ⊂ R′ be an integral ring extension.Moreover, let P,QER be prime ideals with P ⊂ Q, and let P′ER′ be primewith P′∩R = P.

Then there is a prime ideal Q′ER′ with P′ ⊂ Q′ and Q′∩R = Q.

P′ Q′R′: ⊂

R: P ⊂ Q

Proof. As R′ is integral over R, we know that R′/P′ is integral over R/(P′∩R) = R/P by Lemma 9.7(a). Now Q/P is prime by Corollary 2.4, and hence Lying Over as in Proposition 9.18 (b) impliesthat there is a prime ideal in R′/P′ contracting to Q/P, which must be of the form Q′/P′ for a primeideal Q′ER′ with P′ ⊂ Q′ by Lemma 1.21 and Corollary 2.4 again. Now (Q′/P′)∩R/P = Q/Pmeans that Q′∩R = Q, and so the result follows. �

Example 9.25. The ring extension (a) below, which is the same as in Example 9.4 (a), shows atypical case of the Going Up property (note that the correspondence between subvarieties and idealsreverses inclusions, so that the bigger prime ideal Q resp. Q′ corresponds to the smaller subvariety).

R

R′ R′

RP

Q

P′

Q′

QP

P′

(a) (b)

In contrast, in the extension (b) the ring

R′ = C[x,y]/I with I = (xy−1)∩ (x,y)

is the coordinate ring of the variety as in Example 9.4 (b) together with the origin. In this case, itis easily seen geometrically that the extension R ⊂ R′ with R = C[x] satisfies the Lying Over andIncomparability properties as explained in Example 9.19, however not the Going Up property: asin the picture above, the maximal ideal (x,y) of the origin is the only prime ideal in R′ lying overQ, but it does not contain the given prime ideal P′ lying over P. We see from this example thatwe can regard the Going Up property as a statement about the “continuity of fibers”: if we have afamily of points in the base space specializing to a limit point (corresponding to P ⊂ Q in R) anda corresponding family of points in the fibers (corresponding to P′), then these points in the fibershould “converge” to a limit point Q′ over Q. So by Proposition 9.24 the extension (b) above is notintegral (which of course could also be checked directly).

Example 9.26. Let us now consider the opposite “Going Down” direction in Remark 9.23, i. e. thequestion whether we can always construct the smaller prime ideal P′ from Q′ (for given P ⊂ Q ofcourse). Picture (a) below shows this for the extension of Example 9.4 (a).

This time the idea to attack this question is to reduce it to Lying Over by localizing at Q′ instead oftaking quotients, since this keeps exactly the prime ideals contained in Q′ by Example 6.8. Surpris-ingly however, in contrast to Proposition 9.24 the Going Down property does not hold for generalintegral extensions. The pictures (b) and (c) below show typical examples of this (in both cases itcan be checked that the extension is integral):


Q′

glue

R′

P

RR

R′ R′

R

P′

Q′

P

Q′

P QQ Q

(a) (b) (c)

• In case (b) the space corresponding to R′ has two components, of which one is only a singlepoint lying over Q. The consequence is that Going Down obviously fails if we choose Q′

to be this single point. In order to avoid such a situation we can require R′ to be an integraldomain, i. e. to correspond to an irreducible space.

• The case (c) is more subtle: the space R′ = C[x,y] corresponds to a (complex) plane, andgeometrically R is obtained from this by identifying the two dashed lines in the pictureabove. In fact, this is just a 2-dimensional version of the situation of Example 9.11 (b) andExercise 9.12. Although both spaces are irreducible in this case, the singular shape of thebase space corresponding to R makes the Going Down property fail for the choices of P, Q,and Q′ shown above: note that the diagonal line and the two marked points in the top spaceare exactly the inverse image of the curve P in the base. As one might expect from Example9.11 (b), we can avoid this situation by requiring R to be normal.

The resulting proposition is then the following.

Proposition 9.27 (Going Down). Let R ⊂ R′ be an integral ring extension.Assume that R is a normal and R′ an integral domain. Now let P⊂ Q be primeideals in R, and let Q′ER′ be a prime ideal with Q′∩R = Q.

Then there is a prime ideal P′ER′ with P′ ⊂ Q′ and P′∩R = P.

P′ Q′R′: ⊂

R: P ⊂ Q

Proof. The natural map R′→ R′Q′ is injective since R′ is an integral domain. So we can compose itwith the given extension to obtain a ring extension R⊂ R′Q′ as in the picture below on the right. Wewill show that there is a prime ideal in R′Q′ lying over P; by Example 6.8 it must be of the form P′Q′for some prime ideal P′ ⊂ Q′. Since P′Q′ contracts to P′ in R′, we then have P′∩R = P as required.

To prove Lying Over for P in the extension R⊂ R′Q′ it suffices by Proposition9.18 (a) to show that PR′Q′ ∩R⊂ P. So let a∈ PR′Q′ ∩R, in particular a = p

s forsome p ∈ PR′ and s ∈ R′\Q′. We may assume without loss of generality thata 6= 0. As R is normal we can apply Lemma 9.15 (b) to see that the minimalpolynomial of p over K = QuotR is of the form

f = xn + cn−1xn−1 + · · ·+ c0

for some c0, . . . ,cn−1 ∈ P. Note that as a minimal polynomial f is irreducibleover K [G3, Lemma 2.6]. But we also know that a ∈ R ⊂ K, and hence thepolynomial

Q′R′: ⊂

R: P ⊂ Q

R′Q′ : P′Q′

P′

1an f (ax) = xn +

cn−1

axn−1 + · · ·+ c0

an =: xn + c′n−1 xn−1 + · · ·+ c′0

obtained from f by a coordinate transformation is irreducible over K as well. As it is obviouslymonic and satisfies 1

an f (as) = 1an f (p) = 0, it must be the minimal polynomial of s [G3, Lemma

2.6], and so its coefficients c′0, . . . ,c′n−1 lie in R by Proposition 9.15 (a).

90 Andreas Gathmann

Now assume for a contradiction that a /∈ P. The equations c′n−i ai = cn−i ∈ P of elements of R thenimply c′n−i ∈ P for all i = 1, . . . ,n since P is prime. So as s ∈ R′ we see that

sn =−c′n−1 sn−1−·· ·− c′0 ∈ PR′ ⊂ QR′ ⊂ Q′R′ = Q′,

which means that s∈Q′ since Q′ is prime. This contradicts s∈R′\Q′, and thus a∈P as required. �

Exercise 9.28. Let R⊂ R′ be an arbitrary ring extension. Show:

(a) The extension R ⊂ R′ has the Going Up property if and only if for all prime ideals P′ER′

and P = P′∩R the natural map Spec(R′/P′)→ Spec(R/P) is surjective.

(b) The extension R⊂ R′ has the Going Down property if and only if for all prime ideals P′ER′

and P = P′∩R the natural map Spec(R′P′)→ Spec(RP) is surjective.

10. Noether Normalization and Hilbert’s Nullstellensatz 91

10. Noether Normalization and Hilbert’s Nullstellensatz

In the last chapter we have gained much understanding for integral and finite ring extensions. Wenow want to prove an elementary but powerful theorem stating that every finitely generated algebra Rover a field K (so in particular every coordinate ring of a variety by Remark 1.31) is a finite extensionring of a polynomial ring K[z1, . . . ,zr] — and hence of a very simple K-algebra that is easy to dealwith. Let us start by giving the geometric idea behind this so-called Noether Normalization theorem,which is in fact very simple.

Example 10.1 (Idea of Noether Normalization). Let R =C[x1,x2]/(x1x2−1) be the coordinate ringof the variety X =V (x1x2−1)⊂ A2

C as in Example 9.4 (b). We know already that R is not integral(and hence not finite) over C[x1]; this is easily seen geometrically in the picture below on the leftsince this map does not satisfy the Lying Over property for the origin as in Example 9.19.

R

C[y1]

finitey2

2− y21−1 = 0

R

C[x1]

not finitex1x2−1 = 0

coordinatechange

x1 = y2 + y1

x2 = y2− y1

It is easy to change this however by a linear coordinate transformation: if we set e. g. x1 = y2+y1 andx2 = y2−y1 then we can write R also as R =C[y1,y2]/(y2

2−y21−1), and this is now finite over C[y1]

by Proposition 9.5 since the polynomial y22− y2

1− 1 is monic in y2. Geometrically, the coordinatetransformation has tilted the space X as in the picture above on the right so that e. g. the Lying Overproperty now obviously holds. Note that this is not special to the particular transformation that wehave chosen; in fact, “almost any” linear coordinate change would have worked to achieve this goal.

In terms of geometry, we are therefore looking for a change of coordinates so that a suitable coordi-nate projection to some affine space Ar

K then corresponds to a finite ring extension of a polynomialring over K in r variables. Note that this number r can already be thought of as the “dimension” ofX (a concept that we will introduce in Chapter 11) as finite ring extensions correspond to surjectivegeometric maps with finite fibers by Example 9.19, and thus should not change the dimension (wewill prove this in Lemma 11.8).

As we have seen above already, the strategy to achieve our goal is to find a suitable change ofcoordinates so that the given relations among the variables become monic. The first thing we haveto do is therefore to prove that such a change of coordinates is always possible. It turns out that alinear change of coordinates works in general only for infinite fields, whereas for arbitrary fields onehas to allow more general coordinate transformations.

18

Lemma 10.2. Let f ∈ K[x1, . . . ,xn] be a non-zero polynomial over an infinite field K. Assume thatf is homogeneous, i. e. every monomial of f has the same degree (in the sense of Exercise 0.16).

Then there are a1, . . . ,an−1 ∈ K such that f (a1, . . . ,an−1,1) 6= 0.

Proof. We will prove the lemma by induction on n. The case n = 1 is trivial, since a homogeneouspolynomial in one variable is just a constant multiple of a monomial.

92 Andreas Gathmann

So assume now that n> 1, and write f as f =∑di=0 fi xi

1 where the fi ∈K[x2, . . . ,xn] are homogeneousof degree d− i. As f is non-zero, at least one fi has to be non-zero. By induction we can thereforechoose a2, . . . ,an−1 such that fi(a2, . . . ,an−1,1) 6= 0 for this i. But then f ( · ,a2, . . . ,an−1,1) ∈ K[x1]is a non-zero polynomial, so it has only finitely many zeroes. As K is infinite, we can therefore finda1 ∈ K such that f (a1, . . . ,an−1,1) 6= 0. �

Lemma 10.3. Let f ∈K[x1, . . . ,xn] be a non-zero polynomial over an infinite field K. Then there areλ ∈ K and a1, . . . ,an−1 ∈ K such that

λ f (y1 +a1yn,y2 +a2yn, . . . ,yn−1 +an−1yn,yn) ∈ K[y1, . . . ,yn]

is monic in yn (i. e. as an element of R[yn] with R = K[y1, . . . ,yn−1]).

Proof. Let d be the degree of f in the sense of Exercise 0.16, and write f = ∑k1,...,kn ck1,...,knxk11 · · ·xkn

nwith ck1,...,kn ∈ K. Then the leading term of

λ f (y1 +a1yn,y2 +a2yn, . . . ,yn−1 +an−1yn,yn)

= λ ∑k1,...,kn

ck1,...,kn(y1 +a1yn)k1 · · ·(yn−1 +an−1yn)

kn−1yknn

in yn is obtained by always taking the second summand in the brackets and only keeping the degree-dterms, i. e. it is equal to

λ ∑k1,...,kn

k1+···+kn=d

ck1,...,kn ak11 · · ·a

kn−1n−1 yk1+···+kn

n = λ fd(a1, . . . ,an−1,1)ydn ,

where fd is the (homogeneous) degree-d part of f . Now pick a1, . . . ,an−1 by Lemma 10.2 such thatfd(a1, . . . ,an−1,1) 6= 0, and set λ = fd(a1, . . . ,an−1,1)−1. �

Exercise 10.4. Let f ∈ K[x1, . . . ,xn] be a non-zero polynomial over an arbitrary field K. Prove thatthere are λ ∈ K and a1, . . . ,an−1 ∈ N such that

λ f (y1 + ya1n ,y2 + ya2

n , . . . ,yn−1 + yan−1n ,yn) ∈ K[y1, . . . ,yn]

is monic in yn.

Proposition 10.5 (Noether Normalization). Let R be a finitely generated algebra over a field K,with generators x1, . . . ,xn ∈R. Then there is an injective K-algebra homomorphism K[z1, . . . ,zr]→Rfrom a polynomial ring over K to R that makes R into a finite extension ring of K[z1, . . . ,zr].

Moreover, if K is an infinite field the images of z1, . . . ,zr in R can be chosen to be K-linear combina-tions of x1, . . . ,xn.

Proof. We will prove the statement by induction on the number n of generators of R. The case n = 0is trivial, as we can then choose r = 0 as well.

So assume now that n > 0. We have to distinguish two cases:

(a) There is no algebraic relation among the x1, . . . ,xn ∈ R, i. e. there is no non-zero polyno-mial f over K such that f (x1, . . . ,xn) = 0 in R. Then we can choose r = n and the mapK[z1, . . . ,zn]→ R given by zi 7→ xi for all i, which is even an isomorphism in this case.

(b) There is a non-zero polynomial f over K such that f (x1, . . . ,xn) = 0 in R. Then we chooseλ and a1, . . . ,an−1 as in Lemma 10.3 (if K is infinite) or Exercise 10.4 (for any K) and set

y1 := x1−a1xn, . . . , yn−1 := xn−1−an−1xn, yn := xn

(so that x1 = y1 +a1yn, . . . , xn−1 = yn−1 +an−1yn, xn = yn)

or y1 := x1− xa1n , . . . , yn−1 := xn−1− xan−1

n , yn := xn

(so that x1 = y1 + ya1n , . . . , xn−1 = yn−1 + yan−1

n , xn = yn),


respectively. Note that in both cases these relations show that the K-subalgebra K[y1, . . . ,yn]of R generated by y1, . . . ,yn ∈ R is the same as that generated by x1, . . . ,xn, i. e. all of R.Moreover, yn is integral over the K-subalgebra K[y1, . . . ,yn−1] of R, since

λ f (y1 +a1yn, . . . ,yn−1 +an−1yn,yn) or λ f (y1 + ya1n , . . . ,yn−1 + yan−1

n ,yn),

respectively, is monic in yn and equal to λ f (x1, . . . ,xn) = 0. Hence R = K[y1, . . . ,yn] is finiteover K[y1, . . . ,yn−1] by Proposition 9.5. In addition, the subalgebra K[y1, . . . ,yn−1] of R isfinite over a polynomial ring K[z1, . . . ,zr] by the induction hypothesis, and thus R is finiteover K[z1, . . . ,zr] by Lemma 9.6 (a).

Moreover, if K is infinite we can always choose the coordinate transformation of Lemma10.3, and thus y1, . . . ,yn (i. e. also the images of z1, . . . ,zr by induction) are linear combina-tions of x1, . . . ,xn. �

Remark 10.6. Let R = A(X) be the coordinate ring of a variety X over a field K.

(a) In the Noether Normalization of Proposition 10.5, the (images of) z1, . . . ,zr in R are alge-braically independent functions on X in the sense that there is no polynomial relation amongthem with coefficients in K. On the other hand, every other element of R is algebraicallydependent on z1, . . . ,zr, i. e. it satisfies a (monic) polynomial relation with coefficients inK[z1, . . . ,zr]. We can therefore think of r as the “number of parameters” needed to describeX , i. e. as the “dimension” of X as already mentioned in Example 10.1. In fact, we willsee in Remark 11.10 that the number r in Proposition 10.5 is uniquely determined to be thedimension of X in the sense of Chapter 11.

(b) As one would have guessed already from the geometric picture in Example 10.1, the proofof Lemma 10.2 shows that most choices of linear coordinate transformations are suitable toobtain a Noether normalization: in each application of this lemma, only finitely many valuesof a1 ∈ K have to be avoided. Hence we can translate Proposition 10.5 into geometry bysaying that a sufficiently general projection to an r-dimensional linear subspace correspondsto a finite ring extension, and hence to a surjective map with finite fibers (where r is thedimension of X as in (a)).

Exercise 10.7. Find a Noether normalization of the C-algebra C[x,y,z]/(xy+z2,x2y−xy3+z4−1).

Exercise 10.8. Let R ⊂ R′ be an integral ring extension, and assume that R is a finitely generatedalgebra over some field K. Moreover, let P1 ( P3 be prime ideals in R and P′1 ( P′3 be prime ideals inR′ such that P′1∩R = P1 and P′3∩R = P3.

(a) Prove: If there is a prime ideal P2 in R with P1 ( P2 ( P3, thenthere is also a prime ideal P′2 in R′ with P′1 ( P′2 ( P′3.

(b) Can we always find P′2 in (a) such that in addition P′2∩R = P2holds?

R′:

R: ( (P2P1 P3

P′3P′1 P′2( (

As an important application of Noether Normalization we can now give rigorous proofs of somestatements in our dictionary between algebra and geometry, namely of the correspondence between(maximal) ideals in the coordinate ring A(X) of a variety X over an algebraically closed field andsubvarieties (resp. points) of X . There are various related statements along these lines, and they areall known in the literature by the German name Hilbert’s Nullstellensatz (“theorem of the zeroes”).

Let us start with the simplest instance of this family of propositions. Still very algebraic in nature,it is the statement most closely related to Noether Normalization, from which the geometric resultswill then follow easily.

Corollary 10.9 (Hilbert’s Nullstellensatz, version 1). Let K be a field, and let R be a finitelygenerated K-algebra which is also a field.

Then K ⊂ R is a finite field extension. In particular, if in addition K is algebraically closed thenR = K.

94 Andreas Gathmann

Proof. By Noether Normalization as in Proposition 10.5 we know that R is finite over a polynomialring K[z1, . . . ,zr], and thus also integral over K[z1, . . . ,zr] by Proposition 9.5. But R is a field, henceK[z1, . . . ,zr] must be a field as well by Corollary 9.21 (a). This is only the case for r = 0, and so R isfinite over K.

In particular, if K is algebraically closed then there are no algebraic extension fields of K since allzeroes of polynomials over K lie already in K. Hence by Proposition 9.5 there are no finite extensionseither in this case, and we must have R = K. �

Corollary 10.10 (Hilbert’s Nullstellensatz, version 2). Let K be an algebraically closed field. Thenall maximal ideals of the polynomial ring K[x1, . . . ,xn] are of the form

I(a) = (x1−a1, . . . ,xn−an)

for some a = (a1, . . . ,an) ∈ AnK .

Proof. Let PEK[x1, . . . ,xn] be a maximal ideal. Then K[x1, . . . ,xn]/P is a field by Lemma 2.3(b), and in addition certainly a finitely generated K-algebra. Hence K[x1, . . . ,xn]/P = K by Corol-lary 10.9, i. e. the natural map K → K[x1, . . . ,xn]/P, c 7→ c is an isomorphism. Choosing in-verse images a1, . . . ,an of x1, . . . ,xn we get xi = ai for all i, and thus (x1− a1, . . . ,xn− an) ⊂ P.But (x1 − a1, . . . ,xn − an) is a maximal ideal by Example 2.6 (c), and so we must already haveP = (x1−a1, . . . ,xn−an) = I(a). �

Remark 10.11 (Points of a variety correspond to maximal ideals). It is easy to extend Corollary10.10 to a statement about an arbitrary variety X ⊂ An

K over an algebraically closed field K: ifR = A(X) is the coordinate ring of X we claim that there is a one-to-one correspondence

{points of X} 1:1←→ {maximal ideals in A(X)}

a 7−→ I(a)V (I) ←−7 I.

In fact, the maximal ideals of A(X) = K[x1, . . . ,xn]/I(X) are in one-to-one correspondence withmaximal ideals IEK[x1, . . . ,xn] such that I ⊃ I(X) by Lemma 1.21 and Corollary 2.4. By Corollary10.10 this is the same as ideals of the form I(a) = (x1−a1, . . . ,xn−an) containing I(X). But I(a)⊃I(X) is equivalent to a ∈ X by Lemma 0.9 (a) and (c), so the result follows.

Remark 10.12 (Zeroes of ideals in K[x1, . . . ,xn]). Another common reformulation of Hilbert’s Null-stellensatz is that every proper ideal IEK[x1, . . . ,xn] in the polynomial ring over an algebraicallyclosed field K has a zero: by Corollary 2.17 we know that I is contained in a maximal ideal, whichmust be of the form I(a) by Corollary 10.10. But I ⊂ I(a) implies a ∈ V (I) by Lemma 0.9 (a) and(c), and hence V (I) 6= /0.

Note that this statement is clearly false over fields that are not algebraically closed, as e. g. (x2 +1)is a proper ideal in R[x] with empty zero locus in A1

R.19

In order to extend the correspondence between points and maximal ideals to arbitrary subvarietieswe need another algebraic preliminary result first: recall that in any ring R the radical of an ideal Iequals the intersection of all prime ideals containing I by Lemma 2.21. We will show now that it isin fact sufficient to intersect all maximal ideals containing I if R is a finitely generated algebra overa field.

Corollary 10.13 (Hilbert’s Nullstellensatz, version 3). For every ideal I in a finitely generatedalgebra R over a field K we have √

I =⋂

P maximalP⊃I

P.

Proof. The inclusion “⊂” follows immediately from Lemma 2.21, since every maximal ideal isprime.


For the opposite inclusion “⊃”, let f ∈ R with f /∈√

I; we have to find a maximal ideal P ⊃ I withf /∈ P. Consider the multiplicatively closed set S = { f n : n ∈ N}. As f /∈

√I implies I ∩S = /0, we

get by Exercise 6.14 (a) a prime ideal PER with P ⊃ I and P∩ S = /0, in particular with f /∈ P.Moreover, we can assume by Exercise 6.14 (a) that S−1P is maximal. It only remains to show that Pis maximal.

To do this, consider the ring extension K→ R/P→ (R/P) f = R f /Pf , where the subscript f denoteslocalization at S as in Example 6.5 (c). Note that the second map is in fact an inclusion since R/P isan integral domain, and the stated equality holds by Corollary 6.22 (b). Moreover, R f /Pf is a fieldsince Pf is maximal, and finitely generated as a K-algebra (as generators we can choose the classesof generators for R together with 1

f ). So K ⊂ R f /Pf is a finite field extension by Corollary 10.9,and hence integral by Proposition 9.5. But then R/P ⊂ R f /Pf is integral as well, which means byCorollary 9.21 (a) that R/P is a field since R f /Pf is. Hence P is maximal by Lemma 2.3 (b). �

Corollary 10.14 (Hilbert’s Nullstellensatz, version 4). Let X ⊂ AnK be a variety over an alge-

braically closed field K. Then for every ideal IEA(X) we have I(V (I)) =√

I.

In particular, there is a one-to-one correspondence

{subvarieties of X} 1:1←→ {radical ideals in A(X)}

Y 7−→ I(Y )V (I) ←−7 I.

Proof. Let us first prove the equality I(V (I)) =√

I.

“⊂”: Assume that f /∈√

I. By Corollary 10.13 there is then a maximal ideal PEA(X) with P⊃ Iand f /∈ P. But by Remark 10.11 this maximal ideal has to be of the form I(a) = (x1−a1, . . . ,xn−an) for some point a ∈ X . Now I(a)⊃ I implies a ∈V (I) by Lemma 0.9 (a) and(c), and f /∈ I(a) means f (a) 6= 0. Hence f /∈ I(V (I)).

“⊃”: Let f ∈√

I, i. e. f n ∈ I for some n ∈ N. Then ( f (a))n = 0, and hence f (a) = 0, for alla ∈V (I). This means that f ∈ I(V (I)).

The one-to-one correspondence now follows immediately from what we already know: the two mapsare well-defined since I(Y ) is always radical by Remark 1.10, and they are inverse to each other byLemma 0.9 (c) and the statement I(V (I)) =

√I proven above. �

96 Andreas Gathmann

11. Dimension

We have already met several situations in this course in which it seemed to be desirable to have anotion of dimension (of a variety, or more generally of a ring): for example, in the geometric inter-pretation of the Hilbert Basis Theorem in Remark 7.15, or of the Noether Normalization in Remark10.6. We have also used the term “curve” several times already to refer to a “one-dimensional va-riety”, and even if we have not defined this rigorously yet it should be intuitively clear what thismeans. But although this concept of dimension is very intuitive and useful, it is unfortunately alsoone of the most complicated subjects in commutative algebra when it comes to actually provingtheorems. We have now developed enough tools however to be able to study some basic dimensiontheory in this chapter without too many complications. Let us start with the definition of dimension,whose idea is similar to that of the length of a module in Definition 3.18.

Definition 11.1 (Dimension). Let R be a ring.

(a) The (Krull) dimension dimR of R is the maximum number n ∈N such that there is a chainof prime ideals

P0 ( P1 ( · · ·( Pn

of length n in R.

In order to distinguish this notion of dimension from that of a K-vector space V , we willwrite the latter always as dimK V in these notes.

(b) The codimension or height of a prime ideal P in R is the maximum number n ∈N such thatthere is a chain as in (a) with Pn ⊂ P. We denote it by codimR P, or just codimP if the ring isclear from the context.

(c) The dimension dimX of a variety X is defined to be the dimension of its coordinate ringA(X). A 1-dimensional variety is called a curve. The codimension of an irreducible subva-riety Y in X is defined to be the codimension of the prime ideal I(Y ) in A(X) (see Remark2.7 (b)); we denote it by codimX Y or just codimY .

In all cases, we set the dimension resp. codimension formally to ∞ if there is no bound on the lengthof the chains considered. Also, note that in all cases there is at least one such chain (by Corollary2.17 in (a), and the trivial length-0 chain with P0 = P in (b)), hence the definitions above make sense.

Remark 11.2 (Geometric interpretation of dimension). Let X be a variety over an algebraicallyclosed field. By Remark 2.7 (b) and Corollary 10.14, the prime ideals in the coordinate ring A(X) arein one-to-one correspondence with non-empty irreducible subvarieties of X . As this correspondencereverses inclusions, Definition 11.1 says that the dimension of X — or equivalently of its coordinatering A(X) — is equal to the biggest length n of a chain

X0 ) X1 ) · · ·) Xn 6= /0

of irreducible subvarieties of X . Now as in Remark 7.15 the geometric idea behind this definition isthat making an irreducible subvariety smaller is only possible by reducing its dimension, so that ina maximal chain as above the dimension of Xi should be dimX − i, with Xn being a point and X0 anirreducible component of X .

Similarly, the codimension of an irreducible subvariety Y in X is the biggest length n of a chain

X0 ) X1 ) · · ·) Xn ⊃ Y.

Again, for a maximal chain X0 should be an irreducible component of X , and the dimension shoulddrop by 1 in each inclusion in the chain. Moreover, we will have Xn = Y in a maximal chain, so thatwe can think of n as dimX − dimY , and hence as what one would expect geometrically to be thecodimension of Y in X (see Example 11.13 (a)).

11. Dimension 97

Example 11.3.(a) Every field has dimension 0, since the zero ideal is the only prime ideal in this case.(b) More generally, the dimension of a ring R is 0 if and only if there are no strict inclusions

among prime ideals of R, i. e. if and only if all prime ideals are already maximal. So we cane. g. rephrase the theorem of Hopkins in Proposition 7.17 as

R is Artinian ⇔ R is Noetherian and dimR = 0.

Note that this fits well with the geometric interpretation of Remark 11.2, since we havealready seen in Remark 7.15 that Artinian rings correspond to finite unions of points.

(c) Let R be a principal ideal domain which is not a field. Then except for the zero ideal (whichis prime but not maximal) the notions of prime and maximal ideals agree by Example 2.6(b). So the longest chains of prime ideals in R are all of the form 0 ( P for a maximal idealP. It follows that dimR = 1.In particular, the ring Z and polynomial rings K[x] over a field K have dimension 1. Geo-metrically, this means that the variety A1

K (with coordinate ring K[x]) has dimension 1.

Remark 11.4 (Maximal chains of prime ideals can have different lengths). One of the main obstacleswhen dealing with dimensions is that, in general, maximal chains of prime ideals as in Definition11.1 (in the sense that they cannot be extended to a longer chain by inserting more prime ideals) donot necessarily all have the same length. This is easy to see in geometry, where the correspondingstatement is roughly that a space can be made up of components of different dimension. Considere. g. the union X =V (x1x3,x2x3)⊂A3

R of a line and a plane as in Example 0.4 (e), and the followingtwo chains of irreducible subvarieties in X of lengths 1 and 2, respectively.

) )

)

X0

X1

Y0 Y1 Y2

The first chain X0 )X1 is maximal by Example 11.3 (c), since the line has dimension 1. Nevertheless,the second chain is longer (and in fact also maximal, since the plane has dimension 2 as we will seein Proposition 11.9). Hence, due to the two components of X (of different dimension) the maximalchains in X have different lengths. Similarly, the same chains above show that the codimension ofthe point X1 is 1, whereas the codimension of the point Y2 is 2.

Let us state a few properties of dimension that are immediately obvious from the definition.

Remark 11.5 (First properties of dimension). Let R be a ring.

(a) For any prime ideal PER, the prime ideals of R contained in P are in one-to-one correspon-dence with prime ideals in the localization RP by Example 6.8. In other words, we alwayshave codimP = dimRP.

(b) Again let PER be a prime ideal, and set n = dimR/P and m = codimP = dimRP. Then byLemma 1.21 there are chains of prime ideals

P0 ( · · ·( Pm ⊂ P and P⊂ Q0 ( · · ·( Qn

98 Andreas Gathmann

in R that can obviously be glued to a single chain of length m+n. Hence we conclude that

dimR≥ dimR/P+ codimP.

Geometrically, this means for an irreducible subvariety Y of a variety X that

dimX ≥ dimY + codimX Y,

since A(Y ) ∼= A(X)/I(Y ) by Lemma 0.9 (d). Note that we do not have equality in general,since e. g. dimX1 = 0 and codimX1 = 1 in the example of Remark 11.4.

(c) Dimension is a “local concept”: we claim that

dimR = sup{dimRP : P maximal ideal of R}= sup{codimP : P maximal ideal of R}.In fact, if P0 ( · · ·( Pn is a chain of prime ideals in R then by Example 6.8 the correspondinglocalized chain is a chain of primes ideals of the same length in RP, where P is any maximalideal containing Pn. Conversely, any chain of prime ideals in a localization RP correspondsto a chain of prime ideals (contained in P) of the same length in R.Geometrically, we can think of this as the statement that the dimension of a variety is themaximum of the “local dimensions” at every point — so that e. g. the union of a line and aplane in Remark 11.4 has dimension 2.

In the favorable case when all maximal chains of prime ideals do have the same length, the prop-erties of Remark 11.5 hold in a slightly stronger version. We will see in Corollary 11.12 that thisalways happens e. g. for coordinate rings of irreducible varieties. In this case, the dimension andcodimension of a subvariety always add up to the dimension of the ambient variety, and the localdimension is the same at all points.

Lemma 11.6. Let R be a ring of finite dimension in which all maximal chains of prime ideals havethe same length. Moreover, let PER be a prime ideal. Then:

(a) The quotient R/P is also a ring of finite dimension in which all maximal chains of primeideals have the same length;

(b) dimR = dimR/P+ codimP;

(c) dimRP = dimR if P is maximal.

Proof. By Lemma 1.21 and Corollary 2.4, a chain of prime ideals in R/P corresponds to a chainQ0 ( · · ·( Qr of prime ideals in R that contain P. In particular, the length of such chains is boundedby dimR < ∞, and hence dimR/P < ∞ as well. Moreover, if the chain is maximal we must haveQ0 = P, and thus we can extend it to a maximal chain

P0 ( · · ·( Pm = P = Q0 ( · · ·( Qr

of prime ideals in R that includes P. The chains P0 ( · · · ( Pm and Q0 ( · · · ( Qr then mean thatcodimP≥ m and dimR/P≥ r. Moreover, we have m+ r = dimR by assumption. Hence

dimR≥ dimR/P+ codimP≥ dimR/P+m≥ r+m = dimR

by Remark 11.5 (b), and so equality holds. This shows r = dimR/P and hence (a), and of coursealso (b). The statement (c) follows from (b), since for maximal P we have dimR/P = 0 by Example11.3 (a), and dimRP = codimP by Remark 11.5 (a). �

20Exercise 11.7. Let I = Q1∩ ·· ·∩Qn be a primary decomposition of an ideal I in a Noetherian ringR. Show that

dimR/I = max{dimR/P : P is an isolated prime ideal of I}.What is the geometric interpretation of this statement?

Let us now show that, in a Noether normalization K[z1, . . . ,zr]→ R of a finitely generated algebra Rover a field K as in Proposition 10.5, the number r is uniquely determined to be dimR — as alreadymotivated in Remark 10.6. More precisely, this will follow from the following two geometricallyintuitive facts:

11. Dimension 99

• Integral extensions preserve dimension: by Example 9.19, they correspond to surjectivemaps of varieties with finite fibers, and thus the dimension of the source and target of themap should be the same.

• The dimension of the polynomial ring K[x1, . . . ,xn] (and thus of the variety AnK) is n.

We will start with the proof of the first of these statements.

Lemma 11.8 (Invariance of dimension under integral extensions). For any integral ring extensionR⊂ R′ we have dimR = dimR′.

Proof.

“≤” Let P0 ( · · · ( Pn be a chain of prime ideals in R. By Lying Over (for P0) and Going Up(successively for P1, . . . ,Pn) as in Propositions 9.18 and 9.24 we can find a correspondingchain P′0 ( · · · ( P′n of the same length in R′ (where the inclusions have to be strict againsince P′i ∩R = Pi for all i).

“≥” Now let P′0 ( · · ·( P′n be a chain of prime ideals in R′. Intersecting with R we get a chain ofprime ideals P0 ( · · ·( Pn in R by Exercise 2.9 (b), where the inclusions are strict again byIncomparability as in Proposition 9.20. �

For the statement that dimK[x1, . . . ,xn] = n we can actually right away prove the stronger result thatevery chain of prime ideals has length n, and thus e. g. by Lemma 11.6 (c) that the local dimension ofK[x1, . . . ,xn] is the same at all maximal ideals. This is not too surprising if K is algebraically closed,since then all maximal ideals of K[x1, . . . ,xn] are of the form (x1 − a1, . . . ,xn − an) by Corollary10.10, and thus can all be obtained from each other by translations. But for general fields there willbe more maximal ideals in the polynomial ring, and thus the statement that all such localizationshave the same dimension is far less obvious.

Proposition 11.9 (Dimension of polynomial rings). Let K be a field, and let n ∈ N.

(a) dimK[x1, . . . ,xn] = n.

(b) All maximal chains of prime ideals in K[x1, . . . ,xn] have length n.

Proof. We will prove both statements by induction on n, with the case n = 0 being obvious. (In fact,we also know the statement for n = 1 already by Example 11.3 (c)).

So let n ≥ 1, and let P0 ( · · · ( Pm be a chain of prime ideals in K[x1, . . . ,xn]. We have to showthat m ≤ n, and that equality always holds for a maximal chain. By possibly extending the chainwe may assume without loss of generality that P0 = 0, P1 is a minimal non-zero prime ideal, andPm is a maximal ideal. Then P1 = ( f ) for some non-zero polynomial f by Exercise 8.32 (b), sinceK[x1, . . . ,xn] is a unique factorization domain by Remark 8.6.

By a change of coordinates as in the proof of the Noether normalization in Proposition 10.5 wecan also assume without loss of generality that f is monic in xn, and hence that K[x1, . . . ,xn]/P1 =K[x1, . . . ,xn]/( f ) is integral over K[x1, . . . ,xn−1] by Proposition 9.5. We can now transfer our chainof prime ideals from K[x1, . . . ,xn] to K[x1, . . . ,xn−1] as in the diagram below: after dropping the firstprime ideal P0, first extend the chain by the quotient map K[x1, . . . ,xn]→ K[x1, . . . ,xn]/P1, and thencontract it by the integral ring extension K[x1, . . . ,xn−1]→ K[x1, . . . ,xn]/P1.

(· · ·( PmP1K[x1, . . . ,xn]:

P1/P1 · · · ( Pm/P1K[x1, . . . ,xn]/P1: (

(P1/P1)∩K[x1, . . . ,xn−1] ( · · · ( (Pm/P1)∩K[x1, . . . ,xn−1]K[x1, . . . ,xn−1]:

extension

contraction

extension

contraction

100 Andreas Gathmann

We claim that both these steps preserve prime ideals and their strict inclusions, and transfer maximalchains to maximal chains. In fact, for the extension along the quotient map this follows from theone-to-one correspondence between prime ideals in rings and their quotients as in Lemma 1.21 andCorollary 2.4. The contraction then maps prime ideals to prime ideals by Exercise 2.9 (b), andkeeps the strict inclusions by the Incomparability property of Proposition 9.20. Moreover, it alsopreserves maximal chains: in the bottom chain of the above diagram the first entry is 0, the last oneis a maximal ideal by Corollary 9.21 (b), and if we could insert another prime ideal at any step in thechain we could do the same in the middle chain as well by Exercise 10.8 (a).

Now by the induction hypothesis the length m− 1 of the bottom chain is at most n− 1, and equalto n−1 if the chain is maximal. Hence we always have m≤ n, and m = n if the original chain wasmaximal. �

Remark 11.10 (Noether normalization and dimension). Let R be a finitely generated algebra over afield K, and let K[z1, . . . ,zn]→ R be a Noether normalization as in Proposition 10.5. Using Lemma11.8 and Proposition 11.9 we now see rigorously that then the number n is uniquely determinedto be n = dimK[z1, . . . ,zn] = dimR, as already expected in Remark 10.6. In particular, it followsthat a finitely generated algebra over a field (and hence a variety) is always of finite dimension — astatement that is not obvious from the definitions! In fact, the following exercise shows that the finitedimension and Noetherian conditions are unrelated, although rings that meet one of these propertiesbut not the other do not occur very often in practice.

Exercise 11.11 (Finite dimension 6⇐⇒ noetherian). Give an example of a non-Noetherian ring offinite dimension.

In fact, there are also Noetherian rings which do not have finite dimension, but these are hard toconstruct [E, Exercise 9.6].

Corollary 11.12. Let R be a finitely generated algebra over a field K, and assume that R is anintegral domain. Then every maximal chain of prime ideals in R has length dimR.

Proof. By Lemma 1.30 and Lemma 2.3 (a) we can write R = K[x1, . . . ,xn]/P for a prime ideal P ina polynomial ring K[x1, . . . ,xn]. Thus the statement follows from Proposition 11.9 (b) with Lemma11.6 (a). �

Example 11.13. Let Y be an irreducible subvariety of an irreducible variety X . Then by Corollary11.12 every maximal chain of prime ideals in A(X) has the same length, and consequently Lemma11.6 for R = A(X) implies that

(a) dimX = dimY + codimX Y for every irreducible subvariety Y of X (since then P = I(Y ) is aprime ideal and R/P∼= A(Y ));

(b) the local dimension of X is dimX at every point, i. e. all localizations of A(X) at maximalideals have dimension dimX .

Next, let us study how the codimension of a prime ideal is related to the number of generators ofthe ideal. Geometrically, one would expect that an irreducible subvariety given by n equations hascodimension at most n, with equality holding if the equations are “independent” in a suitable sense.More generally, if the zero locus of the given equations is not irreducible, each of its irreduciblecomponents should have codimension at most n.

Algebraically, if I = (a1, . . . ,an) is an ideal in a coordinate ring generated by n elements, the irre-ducible components of V (I) are the maximal irreducible subvarieties of V (I) and thus correspond tothe minimal prime ideals over I as in Exercise 2.23. So our geometric idea above leads us to the ex-pectation that a minimal prime ideal over an ideal generated by n elements should have codimensionat most n.

We will now prove this statement by induction on n for any Noetherian ring. For the proof we needthe following construction of the so-called symbolic powers P(k) of a prime ideal P in a ring R.Their behavior is very similar to that of the ordinary powers Pk; in fact the two notions agree after

11. Dimension 101

localization at P as we will see in the next lemma. The main advantage of the symbolic powers isthat they are always primary (in contrast to the ordinary powers, see Example 8.13).

Lemma 11.14 (Symbolic Powers). Let R be a ring. For a prime ideal PER and n ∈N consider theideal

P(n) := {a ∈ R : ab ∈ Pn for some b ∈ R\P}called the n-th symbolic power of P. Then:

(a) Pn ⊂ P(n) ⊂ P;

(b) P(n) is P-primary;

(c) P(n) RP = Pn RP.

Proof.

(a) The first inclusion is obvious (take b = 1). For the second, let a ∈ P(n), hence ab ∈ Pn ⊂ Pfor some b ∈ R\P. But then a ∈ P since P is prime.

(b) Taking radicals in (a) we see that P =√

P =√

Pn ⊂√

P(n) ⊂√

P = P, so√

P(n) = P. To seethat P(n) is P-primary, let ab ∈ P(n), i. e. abc ∈ Pn for some c ∈ R\P. Then if b /∈

√P(n) = P

we also have bc /∈ P and thus by definition a ∈ P(n). Hence P(n) is P-primary.

(c) For the inclusion “⊂”, let bs ∈P(n) RP, i. e. bc∈Pn for some s,c∈R\P. Then b

s =bcsc ∈Pn RP.

The other inclusions is obvious since P(n) ⊃ Pn. �

Proposition 11.15 (Krull’s Principal Ideal Theorem). Let R be a Noetherian ring, and let a ∈ R.Then every minimal prime ideal P over (a) satisfies codimP≤ 1.

Proof. Let Q′ ⊂Q ( P be a chain of prime ideals in R; we have to prove that Q′ = Q. By the one-to-one correspondence of Example 6.8 between prime ideals in R and in its quotients resp. localizationswe can take the quotient by Q′ and localize at P and prove the statement in the resulting ring, whichwe will again call R for simplicity. Note that this new ring is also still Noetherian by Remark 7.8 (b)and Exercise 7.23. In this new situation we then have:

• by taking the quotient we achieved that Q′ = 0 and R is an integral domain;

• by localizing we achieved that R is local, with unique maximal ideal P;

• we have to prove that Q = 0.

Let us now consider the symbolic powers Q(n) of Lemma 11.14. Obviously, we have Q(n+1) ⊂ Q(n)

for all n. Moreover:

(a) Q(n) ⊂ Q(n+1) +(a) for some n: The ring R/(a) is Noetherian by Remark 7.8 (b) and ofdimension 0 since the unique maximal ideal P/(a) of R/(a) is also minimal by assumption.Hence R/(a) is Artinian by Hopkins as in Example 11.3 (b). This means that the descendingchain

(Q(0)+(a))/(a) ) (Q(1)+(a))/(a) ) · · ·of ideals in R/(a) becomes stationary. Hence Q(n)+(a) = Q(n+1)+(a) for some n, whichimplies Q(n) ⊂ Q(n+1)+(a).

(b) Q(n) = Q(n+1) +PQ(n): The inclusion “⊃” is clear, so let us prove “⊂”. If b ∈ Q(n) thenb = c+ar for some c ∈Q(n+1) and r ∈ R by (a). So ar = b−c ∈Q(n). But a /∈Q (otherwiseP would not be minimal over (a)) and Q(n) is Q-primary by Lemma 11.14 (b), and hencer ∈ Q(n). But this means that b = c+ar ∈ Q(n+1)+PQ(n). 21

Taking the quotient by Q(n+1) in the equation of (b) we see that Q(n)/Q(n+1) = PQ(n)/Q(n+1),and hence that Q(n)/Q(n+1) = 0 by Nakayama’s Lemma as in Exercise 6.16 (a). This meansthat Q(n) = Q(n+1). By Lemma 11.14 (c), localization at Q now gives Qn RQ = Qn+1 RQ, henceQn RQ = (QRQ)Qn RQ by Exercise 1.19 (c), and so Qn RQ = 0 by Exercise 6.16 (a) again. But as Ris an integral domain, this is only possible if Q = 0. �


Exercise 11.16. Let n ∈ N>0, and let P0 ( P1 ( · · ·( Pn be a chain of prime ideals in a Noetherianring R. Moreover, let a ∈ Pn. Prove:

(a) There is a chain of prime ideals P′0 ( P′1 ( · · · ( P′n−1 ( Pn (i. e. in the given chain we maychange all prime ideals except the last one) such that a ∈ P′1.

(b) There is in general no such chain with a ∈ P′0.

Corollary 11.17. Let R be a Noetherian ring, and let a1, . . . ,an ∈ R. Then every minimal prime idealP over (a1, . . . ,an) satisfies codimP≤ n.

Proof. We will prove the statement by induction on n; the case n = 1 is Proposition 11.15. So letn ≥ 2, and let P0 ( · · · ( Ps be a chain of prime ideals in P. After possibly changing some of theseprime ideals (except the last one), we may assume by Exercise 11.16 (a) that an ∈ P1. But then

P1/(an)( · · ·( Ps/(an)

is a chain of prime ideals of length s− 1 in P/(an). As P/(an) is minimal over (a1, . . . ,an−1), wenow know by induction that s−1≤ codimP/(an)≤ n−1, and hence that s≤ n. As the given chainwas arbitrary, this means that codimP≤ n. �

Remark 11.18 (Geometric interpretation of Krull’s Principal Ideal Theorem). Let R be the coordi-nate ring of a variety X ⊂An

K over K, and let I = ( f1, . . . , fr)ER be an ideal generated by r elements.As mentioned above, the minimal prime ideals over I correspond to the irreducible components ofV (I). Hence, Corollary 11.17 states geometrically that every irreducible component of V (I) hascodimension at most r, and thus by Example 11.13 (a) dimension at least n− r.

For the case of a principal ideal (a), we can even give an easy criterion for when equality holds inProposition 11.15. Geometrically, intersecting a variety X with the zero locus of a polynomial freduces the dimension if the polynomial does not vanish identically on any irreducible componentof X , i. e. if f is not a zero divisor in A(X):

Corollary 11.19. Let R be a Noetherian ring, and let a ∈ R not be a zero-divisor. Then for everyminimal prime ideal P over (a) we have codimP = 1.

Proof. Let P1, . . . ,Pn be the minimal prime ideals over the zero ideal as in Corollary 8.30. By Propo-sition 8.27 they can be written as Pi =

√0 : bi for some non-zero b1, . . . ,bn ∈ R.

We claim that a /∈ Pi for all i = 1, . . . ,n. In fact, if a ∈ Pi for some i then a ∈√

0 : bi, and hencear bi = 0 for some r ∈ N>0. But if we choose r minimal then a · (ar−1 bi) = 0 although ar−1 bi 6= 0,i. e. a would have to be a zero-divisor in contradiction to our assumption.

So a /∈ Pi for all i. But on the other hand a ∈ P, and hence P cannot be any of the minimal primeideals P1, . . . ,Pn of R. So P must strictly contain one of the P1, . . . ,Pn, which means that codimP≥ 1.The equality codimP = 1 now follows from Proposition 11.15. �

Example 11.20. Let X = V (x2 + y2− 1) ⊂ A2R be the unit circle. As its coordinate ring is R =

R[x,y]/(x2 + y2−1) and the ideal (x2 + y2−1) is prime, we get as expected that X has dimension

dimX = dimR = dimR[x,y]− codim(x2 + y2−1) (Lemma 11.6 (b) and Proposition 11.9 (b))= 2−1 (Proposition 11.9 (a) and Corollary 11.19)= 1.

With an analogous computation, note that the ring R[x,y]/(x2 + y2 + 1) has dimension 1 as well,although V (x2 + y2 +1) = /0 in A2

R.

Exercise 11.21. Compute the dimension and all maximal ideals of C[x,y,z]/(x2− y2,z2x− z2y).

Exercise 11.22. Show:

(a) Every maximal ideal in the polynomial ring Z[x] is of the form (p, f ) for some prime numberp ∈ Z and a polynomial f ∈ Z[x] whose class in Zp[x] = Z[x]/(p) is irreducible.

11. Dimension 103

(b) dimZ[x] = 2.

Exercise 11.23. For each of the two ring extensions R⊂ R′

(i) R[y]⊂ R[x,y]/(x2− y) (ii) Z⊂ Z[x]/(x2−3)

find:

(a) a non-zero prime ideal PER with a unique prime ideal P′ER′ lying over P, and R′/P′∼=R/P;

(b) a non-zero prime ideal PER with a unique prime ideal P′ER′ lying over P, and R′/P′ 6∼=R/P;

(c) a non-zero prime ideal PER such that there is more than one prime ideal in R′ lying over P.

(In these three cases the prime ideal P is then called ramified, inert, and split, respectively. Can yousee the reason for these names?)

Exercise 11.24. Let R = C[x,y,z,w]/I with I = (x,y)∩ (z,w), and let a = x+ y+ z+w ∈ R. ByKrull’s Principal Ideal Theorem we know that every isolated prime ideal of (a) in R has codimensionat most 1. However, show now that (a) has an embedded prime ideal of codimension 2.

We have now studied the Krull dimension of rings as in Definition 11.1 in some detail. In the restof this chapter, we want to discuss two more approaches to define the notion of dimension. Both ofthem will turn out to give the same result as the Krull dimension in many (but not in all) cases.

The first construction is based on the idea that the dimension of a variety X over a field K can bethought of as the number of independent variables needed to describe its points. In more algebraicterms, this means that dimX should be the maximum number of functions on X that are “alge-braically independent” in the sense that they do not satisfy a polynomial relation with coefficientsin K. This notion of algebraic independence is best-behaved for extensions of fields, and so we willrestrict ourselves to irreducible varieties: in this case A(X) is a finitely generated K-algebra that isan integral domain, and hence we can consider its quotient field QuotR as an extension field of K(i. e. consider rational instead of polynomial functions on X).

Definition 11.25 (Algebraic dependence, transcendence bases). Let K ⊂ L be a field extension, andlet B be a subset of L. As usual, we denote by K(B) the smallest subfield of L containing K and B[G3, Definition 1.13].

(a) The subset B is called algebraically dependent over K if there is a non-zero polynomialf ∈ K[x1, . . . ,xn] with f (b1, . . . ,bn) = 0 for some distinct b1, . . . ,bn ∈ B. Otherwise B iscalled algebraically independent.

(b) The subset B is called a transcendence basis of L over K if it is algebraically independentand L is algebraic over K(B).

Example 11.26.(a) Let K ⊂ L be a field extension, and let a ∈ L. Then {a} is algebraically dependent over K if

and only if a is algebraic over K.

(b) Let R = K[x1, . . . ,xn] be the polynomial ring in n variables over a field K, and let

L = QuotR =

{fg

: f ,g ∈ K[x1, . . . ,xn],g 6= 0}

be its quotient field, i. e. the field of formal rational functions in n variables over K.

This field is usually denoted by K(x1, . . . ,xn), and thus by the same round bracket notationas for the smallest extension field of K containing given elements of a larger field — whichcan be described explicitly as the set of all rational functions in these elements with coeffi-cients in K. Note that this ambiguity is completely analogous to the square bracket notationK[x1, . . . ,xn] which is also used to denote both polynomial expressions in formal variables (toobtain the polynomial ring) or in given elements of an extension ring (to form the subalgebragenerated by these elements).


The set B = {x1, . . . ,xn} is clearly a transcendence basis of L over K: these elements areobviously algebraically independent by definition, and the smallest subfield of L containingK and x1, . . . ,xn is just L itself.

Remark 11.27. In contrast to what one might expect, the definition of a transcendence basis B of afield extension K ⊂ L does not state that L is generated by B in some sense — it is only generated byB up to algebraic extensions. For example, the empty set is a transcendence basis of every algebraicfield extension.

It is true however that a transcendence basis is a “maximal algebraically independent subset”: anyelement a ∈ L\B is algebraic over K(B), and thus B∪{a} is algebraically dependent.

As expected, we can now show that transcendence bases always exist, and that all transcendencebases of a given field extension have the same number of elements.

Lemma 11.28 (Existence of transcendence bases). Let K ⊂ L be a field extension. Moreover, letA⊂ L be algebraically independent over K, and let C ⊂ L be a subset such that L is algebraic overK(C). Then A can be extended by a suitable subset of C to a transcendence basis of L over K.

In particular, every field extension has a transcendence basis.

Proof. Let

M = {B : B is algebraically independent over K with A⊂ B⊂ A∪C}.

Then every totally ordered subset N ⊂ M has an upper bound B: if N = /0 we can take A ∈ M,otherwise

⋃B∈N B. Note that the latter is indeed an element of M:

• Assume that⋃

B∈N B is algebraically dependent. Then f (b1, . . . ,bn) = 0 for a non-zero poly-nomial f over K and distinct elements b1, . . . ,bn ∈

⋃B∈N B. But as N is totally ordered we

must already have b1, . . . ,bn ∈ B for some B ∈ N, in contradiction to B being algebraicallyindependent over K.

• We have A⊂⋃

B∈N B⊂ A∪C since A⊂ B⊂ A∪C for all B ∈ N.

Hence M has a maximal element B. We claim that B is a transcendence basis of L over K. In fact, asan element of M we know that B is algebraically independent. Moreover, since B is maximal everyelement of C is algebraic over K(B), hence K(C) is algebraic over K(B) [G3, Exercise 2.27 (a)], andso L is algebraic over K(B) by Lemma 9.6 (b) since L is algebraic over K(C) by assumption. �

Proposition and Definition 11.29 (Transcendence degree). Let K ⊂ L be a field extension. If L hasa finite transcendence basis over K, then all such transcendence bases are finite and have the samenumber n of elements. In this case we call this number n the transcendence degree trdegK L of Lover K. Otherwise, we set formally trdegK L = ∞.

Proof. Let B and C be transcendence bases of L over K, and assume that B = {b1, . . . ,bn} has nelements. By symmetry, it suffices to show that |C| ≤ |B|. We will prove this by induction on n, withthe case n = 0 being obvious since then K ⊂ L is algebraic, and hence necessarily C = /0.

So assume now that n > 0. As we are done if C ⊂ B, we can pick an element c ∈ C\B, whichis necessarily transcendental. By Lemma 11.28 we can extend it to a transcendence basis B′ byelements of B. Note that B′ can have at most n elements: the set {c,b1, . . . ,bn} is algebraicallydependent since B is a transcendence basis.

So B′ and C are two transcendence bases of L over K that both contain c. Hence B′\{c} and C\{c}are two transcendence bases of L over K(c). By induction this means that |B′\{c}| = |C\{c}|, andhence that |C|= |C\{c}|+1 = |B′\{c}|+1≤ n = |B| as desired. �

Example 11.30.(a) A field extension is algebraic if and only if it has transcendence degree 0.

11. Dimension 105

(b) Given that π is transcendental over Q, the field extensions Q ⊂ Q(π) and Q ⊂ Q(π,√

2)both have transcendence degree 1, with {π} as a transcendence basis. The field extensionQ ⊂ C has infinite transcendence degree, since the set of all rational functions in finitelymany elements with coefficients in Q is countable, whereas C is uncountable.

(c) The quotient field K(x1, . . . ,xn) of the polynomial ring K[x1, . . . ,xn] has transcendence degreen over K, since {x1, . . . ,xn} is a transcendence basis by Example 11.26 (b).

With the notion of transcendence degree we can now give another interpretation of the dimensionof an irreducible variety — or more generally of a finitely generated algebra over a field which is adomain.

Proposition 11.31 (Transcendence degree as dimension). Let R be a finitely generated algebra overa field K, and assume that R is an integral domain. Then dimR = trdegK QuotR.

Proof. Let K[z1, . . . ,zn]→ R be a Noether normalization as in Proposition 10.5, where n = dimRby Remark 11.10. Then QuotR is algebraic over QuotK[z1, . . . ,zn] = K(z1, . . . ,zn): if a,b ∈ R withb 6= 0 then a and b are integral over K[z1, . . . ,zn], hence algebraic over K(z1, . . . ,zn). ThereforeK(z1, . . . ,zn)(a,b) is algebraic over K(z1, . . . ,zn) [G3, Exercise 2.27 (a)], which implies that a

b ∈QuotR is algebraic over K(z1, . . . ,zn).

But this means that the transcendence basis {z1, . . . ,zn} of K(z1, . . . ,zn) over K (see Example 11.26(b)) is also a transcendence basis of QuotR over K, i. e. that trdegK QuotR = n = dimR. �

Exercise 11.32. Let K be a field, and let R be a K-algebra which is an integral domain. If R is finitelygenerated we know by Proposition 11.31 that dimR = trdegK QuotR. Now drop the assumption thatR is finitely generated and show:

(a) dimR≤ trdegK QuotR.

(b) The inequality in (a) may be strict.

Exercise 11.33. Let X ⊂ AnK and Y ⊂ Am

K be varieties over a field K. Prove:

(a) dim(X×Y ) = dimX +dimY .

(b) If n=m then every irreducible component of X∩Y has dimension at least dimX +dimY−n.22

Finally, another approach to the notion of dimension is given by linearization. If a is a point of avariety X ⊂An

K over a field K we can try to approximate X around a by an affine linear space in AnK ,

and consider its dimension to be some sort of local dimension of X at a. Let us describe the preciseconstruction of this linear space.

Construction 11.34 (Tangent spaces). Let us consider a variety X ⊂ AnK with coordinate ring

A(X) = K[x1, . . . ,xn]/I(X), and let a = (a1, . . . ,an) ∈ X . By a linear change of coordinatesyi := xi−ai we can shift a to the origin.

Consider the “differential” map

d : K[x1, . . . ,xn]→ K[y1, . . . ,yn], f 7→n

∑i=1

∂ f∂xi

(a) · yi,

where ∂ f∂xi

denotes the formal derivative of f with respect to xi [G1, Exer-cise 9.10]. It can be thought of as assigning to a polynomial f its linearterm in the Taylor expansion at the point a. We then define the tangentspace to X at a to be

TaX :=V ({d f : f ∈ I(X)}) ⊂ Kn, (∗)

X

a

TaX

i. e. we take all equations describing X , linearize them at the point a, and take the common zero locusof these linearizations. As in the picture above we can therefore think of TaX as the linear space that


gives the best approximation of X at a. Note that it suffices in (∗) to let f range over a generatingset for I(X), since for f ,g ∈ I(X) and h ∈ K[x1, . . . ,xn] we have

d( f +g) = d f +dg and d(h f ) = h(a)d f + f (a)dh = h(a)d f .

In order to transfer these ideas to other non-geometric cases we need an alternative description of thetangent space: let P = I(a) = (y1, . . . ,yn)EA(X) be the (maximal) ideal of the point a, and considerthe K-linear map

ϕ : P→ HomK(TaX ,K), f 7→ d f |TaX

(that is well-defined by the definition (∗) of TaX). It is clearly surjective since every linear map onTaX must be a linear combination of the coordinate functions. Moreover, we claim that kerϕ = P2:

“⊂” Let ϕ( f ) = d f |TaX = 0. Note that L := {dg : g ∈ I(X)} is a linear subspace of 〈y1, . . . ,yn 〉of some dimension k. Its zero locus TaX then has dimension n− k. Hence the space of alllinear forms vanishing on TaX has dimension k again and clearly contains L, and thus mustbe equal to L. As d f lies in this space, we conclude that d f = dg for some g ∈ I(X). Thend( f −g) = 0, and thus the Taylor expansion of f −g at a does not contain constant or linearterms. Hence f = f −g ∈ P2.

“⊃” The K-vector space P2 is generated by products f g for f ,g ∈ P = I(a), and we clearly haveϕ( f g) = f (a)dg|TaX +g(a)d f |TaX = 0 since f (a) = g(a) = 0.

So by the homomorphism theorem we get a natural isomorphism P/P2 → HomK(TaX ,K) of K-vector spaces. In other words, the tangent space TaX is naturally the dual vector space of P/P2. Inparticular, its dimension is dimK TaX = dimK P/P2.

Remark 11.35 (Tangent spaces are local). Let P be a maximal ideal in a ring R, and let K = R/P.Then the R-module P/P2 is also a K-vector space (with the same multiplication), and the classes ofelements of S = R\P in K are invertible. Hence localizing P/P2 at S is an isomorphism of K-vectorspaces, i. e. we get

P/P2 ∼= S−1(P/P2) ∼= S−1P/S−1P2

by Corollary 6.22 (b). So the tangent space TaX of a variety X at a point a ∈ X as in Construction11.34 can equally well be obtained from the local ring of X at a.

This observation allows us to assign to any local ring R (with maximal ideal P and K = R/P) anumber dimK P/P2 that can be interpreted as the dimension of the tangent space if R is the ring oflocal functions on a variety at a point. Let us see how this number compares to the actual dimensiondimR of R. For simplicity, we will restrict our attention to Noetherian rings.

Lemma 11.36. Let R be a local Noetherian ring with maximal ideal P, and set K = R/P.

(a) The number dimK P/P2 is the minimum number of generators for the ideal P.

(b) dimR≤ dimK P/P2 < ∞.

Proof. Of course, the ideal P is finitely generated since R is Noetherian. Let n be the minimumnumber of generators for P.

(a) Let P = (a1, . . . ,an). Then a1, . . . ,an generate P/P2 as an R/P-vector space, and thusdimK P/P2 ≤ n. Moreover, if a1, . . . ,an were linearly dependent then after relabelinga1, . . . ,an−1 would still generate P/P2 — but then Nakayama’s Lemma as in Exercise 6.16(b) (for M = I = P) implies that a1, . . . ,an−1 generate P as an R-module, in contradiction tothe minimality of n. Hence dimK P/P2 = n.

(b) By Remark 11.5 (c), Remark 11.5 (a), and Corollary 11.17 we have dimR = dimRP =codimP≤ n < ∞. Hence the result follows since n = dimK P/P2 by (a). �

Example 11.37. Consider the curves X1 and X2 in A2R with ideals I(X1) = (x2− x2

1) and I(X2) =

(x22− x2

1− x31), respectively, as in the picture below.

11. Dimension 107

T0X1

T0X2

x1

x2 x2

x1

X1 =V (x2− x21) X2 =V (x2

2− x21− x3

1)

By Construction 11.34, their tangent spaces T0X1 and T0X2 at the origin are the zero loci of the linearparts of their defining equations, i. e.

T0X1 =V (x2) is the x1-axis, and

T0X2 =V (0) is all of R2.

Hence for X1, the tangent space T0X1 has the same dimension as X1, i. e. the first inequality inLemma 11.36 (b) is an equality. Geometrically, this means that X1 has a “good” approximation at 0by a linear space of the same dimension — which is the usual understanding of a tangent space. Inthe case of X2 however, there is no reasonable approximation at 0 by a linear space of dimension 1.Consequently, the dimension of the tangent space is bigger than that of X , i. e. the first inequality ofLemma 11.36 (b) is strict. This is usually expressed by saying that the origin is a singular point ofX2, whereas it is a regular or smooth point of X1. The precise definition of these terms is as follows.

Definition 11.38 (Regular local rings, regular points of a variety).

(a) Let R be a local ring with maximal ideal P, and set K = R/P. We say that R is regular if itis Noetherian and dimR = dimK P/P2.

(b) Let X be a variety. A point a ∈ X is called regular or smooth if its ring of local functionsA(X)I(a) as in Example 6.5 (d) is regular, i. e. by Construction 11.34 if dimX = dimTaX(note that A(X)I(a) is always Noetherian by Remark 7.15 and Exercise 7.23). Otherwise wesay that a is a singular point of X .

Example 11.39.

(a) Any field is a regular local ring (of dimension 0).

(b) Let X ⊂ AnK be a variety over a field K, with ideal I(X) = ( f1, . . . , fr)EK[x1, . . . ,xn]. By

Construction 11.34, the tangent space TaX to X at a point a ∈ X is the kernel of the Jacobianmatrix

J f (a) :=(

∂ fi

∂x j(a))

1≤i≤r1≤ j≤n

.

Hence a is a regular point of X if and only if dimX = dimK kerJ f (a). In particular, this isthe case if J f (a) has maximal row rank r, because then

dimX ≤ dimK kerJ f (a) (Lemma 11.36 (b))= n− r

≤ dimX ,

since by Remark 11.18 every irreducible component of X = V ( f1, . . . , fr) has dimension atleast n− r. Note that this fits well with the implicit function theorem [G2, Proposition 27.9]in analysis, which states that (in the case K = R) a Jacobi matrix with maximal row rank ata point a guarantees that X is locally the graph of a continuously differentiable function, andthus that X is regular at a in the sense of Example 11.37.

Explicitly, the standard parabola X1 in Example 11.37 is regular at every point since itsJacobian matrix (−2x1 1) has full row rank everywhere, whereas the curve X2 has a singularpoint at the origin as its Jacobian matrix (−2x1−3x2

1 2x2) vanishes there.


Finally, we will prove the important result that any regular local ring is an integral domain — astatement that is easy to understand geometrically since a variety that is regular at a point in thesense of Example 11.37 should not consist of several components that meet in this point.

Proposition 11.40. Any regular local ring is an integral domain.

Proof. Let R be a regular local ring of dimension n with maximal ideal P, and set K = R/P. We willprove the statement of the proposition by induction on n. For n = 0 we have dimK P/P2 = 0, henceP = P2, so P = 0 by Nakayama’s Lemma as in Exercise 6.16 (a). But if the zero ideal is maximalthen R is a field, and hence obviously an integral domain.

So let now n > 0, and let P1, . . . ,Pr be the minimal primes over the zero ideal as in Corollary 8.30.Note that P 6⊂ P2∪P1∪ ·· ·∪Pr since otherwise by Exercise 2.10 (c) we would have P ⊂ P2 (henceP = P2, and thus n = 0 by Lemma 11.36 (b)) or P ⊂ Pi for some i (hence Pj ⊂ P ⊂ Pi for all j, sor = 1 and P is the unique maximal and the unique minimal prime, which means again that n = 0).We can therefore find an element a ∈ P with a /∈ P2 and a /∈ Pi for all i.

Consider the ring R/(a), which is again local (with maximal ideal P/(a), and (P/(a))/(P/(a))2 ∼=P/(P2 +(a))) and Noetherian (by Remark 7.8 (b)). We will show that it is regular of dimensionn−1:

• As a /∈ P2, we can extend the element a ∈ P/P2 to a basis (a,a2, . . . ,an) of P/P2. Thena2, . . . ,an generate the vector space P/(P2 +(a)), and hence dimP/(P2 +(a))≤ n−1.

• Let Pi = Q0 ( · · ·( Qn = P be a maximal chain of prime ideals in R; note that it has to startwith a minimal prime ideal Pi and end with P. As a ∈ P, we can arrange this by Exercise11.16 (a) so that a ∈ Q1. Then Q1/(a) ( · · · ( Qn/(a) is a chain of prime ideals of lengthn−1 in R/(a), and so dimR/(a)≥ n−1.

Together with Lemma 11.36 (b) we conclude that

n−1≤ dimR/(a)≤ dimP/(P2 +(a))≤ n−1,

and hence that R/(a) is regular of dimension n−1.

By induction, this means that R/(a) is an integral domain, and hence by Lemma 2.3 (a) that (a) isa prime ideal. Hence we must have Pi ⊂ (a) for some i. This means for any b ∈ Pi that b = ac forsome c ∈ R, hence c ∈ Pi since a /∈ Pi and Pi is prime, and therefore b ∈ P ·Pi. In total, this meansthat Pi = P ·Pi, and so Pi = 0 by Nakayama as in Exercise 6.16 (a). Thus the zero ideal is prime, i. e.R is an integral domain. �

23

12. Valuation Rings 109

12. Valuation Rings

In the remaining two chapters of this class we will restrict our attention to 1-dimensional rings,i. e. geometrically to curves. More precisely, we will only deal with 1-dimensional rings whoselocalizations at all maximal ideals are regular. For a curve X over an algebraically closed field, whenthe maximal ideals of A(X) are in one-to-one correspondence with points of X by Remark 10.11,this means precisely that we will require X to be smooth, i. e. that all points of X are smooth in thesense of Example 11.37 and Definition 11.38.

In the current chapter we will consider such rings and varieties from a local point of view, postponingtheir global study to Chapter 13. Such 1-dimensional regular local rings are probably the “nicest”rings (that are not fields) — they have a very special structure as we will see in the following keylemma and its geometric interpretation.

Lemma 12.1. Let R be a 1-dimensional regular local ring.

(a) The maximal ideal P of R is a non-zero principal ideal.

(b) For every element a ∈ R\{0} there is a unique number n ∈ N such that (a) = Pn. We willcall it the valuation of a (see Remark 12.15 (a)).

Proof.

(a) By assumption, the vector space dimension of P/P2 over R/P is 1. But this is also theminimum number of generators for P by Lemma 11.36 (a), hence P is principal and non-zero.

(b) By Proposition 11.40 we know that R is an integral domain. So it has a unique minimalprime ideal 0 and a unique maximal ideal P, and since dimR = 1 these two prime ideals arein fact the only ones. Hence by Lemma 2.21 we know that

√(a) is equal to P or R. In both

cases it follows that Pn ⊂ (a) for some n ∈ N by Exercise 7.22 (b). Since P is principal by(a), we can write this equivalently as tn = ba for some b ∈ R and a generator t of P.

Now choose n minimal with this property. Then b must be a unit: otherwise (b) is containedin a maximal ideal by Corollary 2.17, which must be P. Hence b = b′t for some b′ ∈ R,so tn = b′ta, and thus tn−1 = b′a since R is an integral domain — in contradiction to theminimality of n.

So with this choice of n we get tn = ba for a unit b, which means that (a) = (tn) = Pn.Moreover, no other choice of exponent would work: if we had tk = ca for a unit c andk < n (resp. k > n), then this would imply that tn−k = bc−1 (resp. tk−n = cb−1) is a unit, incontradiction to t ∈ P. �

Remark 12.2 (Geometric interpretation: orders of vanishing). Let x be afixed smooth point on a curve X , and let R be the ring of local functions onX at x, i. e. the localization of A(X) at the maximal ideal I(x) as in Example6.5 (d). Then R is a 1-dimensional regular local ring.

Xx

The maximal ideal P of R then consists of all local functions that vanish at x, and a generator t for Pas in Lemma 12.1 (a) can be thought of as a local coordinate for X in a neighborhood of x (that hasthe value 0 at x, and vanishes to order 1 there). Consequently, if a ∈ R is a local function on X at xthe equation (a) = Pn of Lemma 12.1 (b), i. e. a = ctn for a unit c, means that a vanishes to order nat x. We can therefore think of the valuation constructed in Lemma 12.1 as an order of vanishing ofa local function on a curve at a point.


Actually, the full algebraic notion of a valuation is more general than above: one considers valuationson general integral domains and in fact also on their quotient fields, and allows their values to be inany ordered Abelian group instead of just in N or Z. Let us give the precise definition now, returningto our special case of 1-dimensional regular local rings later in Proposition 12.14.

Definition 12.3 (Valuations and valuation rings).(a) An ordered group is an Abelian group G (usually written additively) with a total order ≤

such that m≤ n implies m+ k ≤ n+ k for all m,n,k ∈ G.

(b) Let K be a field; as usual we write K∗ for K\{0}. A valuation on K is a map ν : K∗→G foran ordered group G such that for all a,b ∈ K∗ we have

(i) ν(ab) = ν(a)+ν(b) (i. e. ν is a homomorphism of groups); and

(ii) ν(a+b)≥min(ν(a),ν(b)) if a+b 6= 0.

We extend this map formally to all of K by setting ν(0) := ∞, so that (i) and (ii) hold for alla,b ∈ K.

(c) For a valuation ν : K∗→ G the subgroup ν(K∗)≤ G is called the value group, and

Rν := {a ∈ K : ν(a)≥ 0}the valuation ring of ν . Note that it is indeed a ring: we clearly have ν(0) = ∞ ≥ 0 andν(1) = 0, and for a,b ∈ K with ν(a) ≥ 0 and ν(b) ≥ 0 it follows immediately from (i) and(ii) above that ν(a+b)≥ 0 and ν(ab)≥ 0 as well.

Example 12.4.(a) Any field K allows the trivial valuation ν : K∗→{0}, a 7→ 0. Its value group is {0}, and its

valuation ring is K itself. Note that even in this case we still have ν(0) = ∞ by definition.

(b) The groups Z, Q, and R are all ordered. In fact, the most common non-trivial value groupin our examples will be Z (see Proposition 12.13). But for general valuations other groupsmay occur as well, as we will see e. g. in (e) below, and in the proof of Proposition 12.8.

(c) The most important example of a valuation is the following. Fix a prime element p in aunique factorization domain R, and let K = QuotR. Note that every non-zero element of Rcan be written uniquely as a pn for some n ∈N and a ∈ R with p 6 | a, and consequently everyelement of K∗ can be written uniquely as a pn for some n ∈ Z and a ∈ K∗ that can be writtenas a quotient of elements of R that do not contain p as a factor. With this representation thereis an obvious map

ν : K∗→ Z, a pn 7→ nthat assigns to every element the exponent of p in its prime factorization. It is clearly well-defined and a homomorphism of groups. It also satisfies condition (ii) of Definition 12.3 (b):for any two such elements a pn and b pm (with a,b ∈ K∗ not containing the prime factor p,and n,m ∈ Z), without loss of generality with n≥ m, we have

ν(a pn +b pm) = ν(

pm (a pn−m +b))= m+ν(a pn−m +b)

(∗)≥ m = ν(b pm),

where (∗) follows since a pn−m +b does not contain the factor p in its denominator. Henceν is a valuation on K. Its value group is obviously Z, and its valuation ring is

Rν = {a pn : a ∈ K∗, n ∈ N, a does not contain p}∪{0},i. e. precisely the localization R(p) of R at the prime ideal (p) as in Example 6.5 (d).

(d) Let L be a field, and let

K =

{∑n∈Z

antn : an ∈ L for all n, the set {n ∈ Z : an 6= 0} is bounded below}

be the set of all formal “power series with integer exponents” in one variable t. As in the caseof the usual power series ring L[[t]], there are no convergence requirements on the coefficientsof the series [G1, Definition 9.1]. We do require however that for a non-zero element f ∈


K there is a minimum exponent of t occurring in f , which implies that we can write funiquely as f = tn g for some n ∈ Z and g ∈ L[[t]] with non-zero constant coefficient. Withthis definition K is in fact a field: it is obvious that it is a ring, and since a power series withnon-zero constant coefficient is invertible [G1, Exercise 9.12], a multiplicative inverse of tn gas above is t−n g−1. We call K the field of (formal) Laurent series over L.

It is now obvious that

ν : K∗→ Z, ∑n∈Z

antn 7→min{n ∈ Z : an 6= 0}

is a valuation on K. Its valuation ring is simply the ring L[[t]] of power series over L. In fact,this construction is a special case of (c) since one can show that L[[t]] is a unique factorizationdomain with t ∈ L[[t]] prime, and K = Quot(L[[t]]).

(e) The idea of (d) can be generalized to construct a valuation with value group Q: for a field Lwe now consider the set

K =

{∑

q∈Qaqtq : aq ∈ L for all q, the set {q ∈Q : aq 6= 0} is bounded below

and has bounded denominators}

of all formal “power series with rational exponents”, and the same map

ν : K∗→Q, ∑q∈Q

aqtq 7→min{q ∈Q : aq 6= 0}

as above. It can then be checked that K is again a field — we will not do this here as wewill not need this field again — and that ν is a valuation with value group Q. The field K iscalled the field of (formal) Puiseux series over L.

Remark 12.5.(a) We will see in Propositions 12.13 and 12.14 that a 1-dimensional regular local ring is a

unique factorization domain, and that in this case the construction of Example 12.4 (c) spe-cializes to our original situation of Lemma 12.1 and Remark 12.2. For the moment it sufficesto note that, in the geometric situation, we should imagine K in Definition 12.3 to be the fieldof rational functions on a variety X around a smooth point a. The valuation ν( f ) ∈ Z of anelement f ∈ K∗ can then be thought of as the order of the zero (if ν( f ) > 0) or pole (ifν( f )< 0) of f at a ∈ X .

(b) It is clear from Definition 12.3 that every valuation ring Rν of a valuation ν is an integral do-main, since it is a subring of a field. In fact, valuation rings have many more nice properties.Let us prove the most important ones now.

Lemma 12.6 (Properties of valuation rings). Let R = Rν be the valuation ring of a valuation ν :K∗→ G as in Definition 12.3.

(a) For all a ∈ K∗ we have a ∈ R or a−1 ∈ R.

(b) For all a,b ∈ R we have ν(a)≤ ν(b) if and only if b ∈ (a).

(c) The group of units of R is R∗ = {a ∈ R : ν(a) = 0}.(d) R is a local ring with maximal ideal P = {a ∈ R : ν(a)> 0}.(e) R is a normal domain.

Proof.

(a) Since ν(a)+ν(a−1) = ν(aa−1) = ν(1) = 0 we must have ν(a)≥ 0 or ν(a−1)≥ 0, and thusa ∈ R or a−1 ∈ R.

(b) The statement is obvious for a = 0, so let us assume that a 6= 0. If ν(a)≤ ν(b) then ν( ba ) =

ν(b)−ν(a) ≥ 0, hence ba ∈ R, and thus b = b

a · a ∈ (a). Conversely, if b ∈ (a) then b = cafor some c ∈ R, and therefore ν(b) = ν(c)+ν(a)≥ ν(a).


(c) Let a∈R with a 6= 0, so ν(a)≥ 0. Then a∈R∗ if and only if a−1 ∈R, i. e. ν(a−1) =−ν(a)≥0 as well, which is the case if and only if ν(a) = 0.

(d) It is clear that P is an ideal, since a,b ∈ P and r ∈ R imply ν(a+b)≥min(ν(a),ν(b))> 0and ν(ra) = ν(r)+ν(a)> 0. Moreover, any bigger ideal contains an element with valuation0, i. e. a unit by (c), and thus is equal to R.

(e) Let a ∈ K∗ be integral over R. Then an + cn−1an−1 + · · ·+ c0 = 0 for some n ∈ N andc0, . . . ,cn−1 ∈ R. Assume that a /∈ R. Then a−1 ∈ R by (a), which leads to the contradiction

a =−cn−1−·· ·− c0 a−n+1 ∈ R.

Hence we must have a ∈ R as claimed. �

So far, we have always started with a valuation ν on a field K, and then constructed a valuation ringRν ⊂ K from it. We will now show that this process can be reversed: the valuation ring Rν itselfcontains in fact enough information to reconstruct the valuation ν from it.

Construction 12.7 (Reconstruction of the valuation from its valuation ring). Let R = Rν be thevaluation ring of a valuation ν : K∗→ G as in Definition 12.3. Then ν can be recovered completelyfrom R up to isomorphisms in the following sense:

(a) We must have K = QuotR: the inclusion “⊃” is obvious, and “⊂” follows from Lemma 12.6(a), since both a ∈ R and a−1 ∈ R imply a ∈ QuotR.

(b) The group homomorphism ν : K∗→ G has kernel R∗ by Lemma 12.6 (c), and image ν(K∗).Hence the homomorphism theorem implies that the value group ν(K∗) is isomorphic toK∗/R∗ (which is uniquely determined by R due to (a)), and that with this isomorphism thevaluation map ν : K∗→ K∗/R∗ is just the quotient map.

(c) The order on ν(K∗) is determined by R since for a,b ∈ K∗ we have ν(a)≤ ν(b) if and onlyif ν( b

a )≥ 0, i. e. if ba ∈ R.

So we can say that every valuation ring belongs to a unique valuation (if we assume the valuationto be surjective, i. e. we can of course not recover the whole group G from R, but only the valuationsubgroup ν(K∗) of G). In fact, there is even an easy criterion to determine whether a given ring issuch a valuation ring:

Proposition 12.8 (Alternative characterization of valuation rings). For a ring R the following state-ments are equivalent:

(a) R is the valuation ring of a valuation ν : K∗→ G (which is then unique up to isomorphismsby Construction 12.7).

(b) R is an integral domain, and for all a ∈ QuotR\{0} we have a ∈ R or a−1 ∈ R.

Proof. The implication (a) ⇒ (b) is just Lemma 12.6 (a). For the opposite direction, let R be anintegral domain, set K = Quot(R) and G = K∗/R∗. Note that, in contrast to Definition 12.3 (a), thegroup G will be written multiplicatively in this proof. Now

a≤ b :⇔ ba∈ R

is a well-defined relation on G (for a,b ∈ K∗ and c,d ∈ R∗ with ba ∈ R we have bd

ac ∈ R) and makes Ginto an ordered group (for a,b,c ∈ K∗ with b

a ∈ R we have bcac ∈ R). Let ν : K∗→ G be the quotient

map, so that ν(a)≤ ν(b) for a,b ∈ K∗ if and only if ba ∈ R. Then ν is a valuation:

(i) The relation ν(ab) = ν(a)ν(b) is obvious.

(ii) Let a,b ∈ K∗. By assumption we have ab ∈ R or b

a ∈ R. In the first case a+bb = a

b +1 ∈ R andthus ν(a+b)≥ ν(b), whereas in the second case we get similarly ν(a+b)≥ ν(a). Henceν(a+b)≥min(ν(a),ν(b)).


Its valuation ring Rν is now exactly R, since we have ν(a) ≥ 0 = ν(1) for a ∈ K∗ if and only ifa1 ∈ R. �

Remark 12.9. Proposition 12.8 states that we could alternatively define a valuation ring to be anintegral domain R such that a ∈ R or a−1 ∈ R for all a ∈ QuotR\{0}. In fact, one will often findthis definition in the literature, and in the following we will also often talk about a valuation ring Rwithout mentioning a valuation first. Our results above show that even then the ring R always has aunique valuation ν : K→ G associated to it such that R = Rν and ν is surjective.

24This alternative characterization of valuation rings allows us to prove some more of their propertiesthat would be harder to see using the original Definition 12.3. The following lemma and propositiongive two examples of this.

Lemma 12.10 (Enlarging valuation rings). Let R be valuation ring, and let R′ be another ring withR⊂ R′ ⊂ QuotR. Then R′ is also a valuation ring, with QuotR′ = QuotR.

Proof. Taking quotient fields in the inclusion R ⊂ R′ ⊂ QuotR we see that QuotR′ = QuotR. Nowif a ∈ QuotR′ = QuotR we have a ∈ R or a−1 ∈ R by Proposition 12.8, and hence also a ∈ R′ ora−1 ∈ R′. Therefore R′ is a valuation ring by Proposition 12.8 again. �

Proposition 12.11 (Integral closure from valuation rings). Let R be an integral domain, and let Rbe its integral closure in QuotR. Then

R =⋂

R⊂R′⊂QuotRR′ valuation ring

R′.

Proof.

“⊂” Let x ∈ R, and let R′ be a valuation ring with R ⊂ R′ ⊂ QuotR. So x ∈ QuotR = QuotR′ isintegral over R, and thus also over R′. But then x ∈ R′ since R′ is normal by Lemma 12.6 (e).

“⊃” Let x /∈ R. We will construct a valuation ring R′ with R⊂ R′ ⊂ QuotR and x /∈ R′.

Note that x /∈ R[ 1x ], since otherwise there would be a relation x = a0 + a1x−1 + · · ·+ anx−k

with a0, . . . ,ak ∈ R, which means that xk+1− a0xk− a1xk−1− ·· ·− ak = 0, and thus that xwould be integral over R in contradiction to x /∈ R. So the set

M = {R′ : R′ is a ring with R[ 1x ]⊂ R′ ⊂ QuotR and x /∈ R′}

is non-empty since it contains R[ 1x ]. By Zorn’s Lemma as in Proposition 2.16 it contains a

maximal element R′. Of course we then have R⊂ R′ ⊂QuotR and x /∈ R′, so it only remainsto show that R′ is a valuation ring.

We will prove this using the criterion of Proposition 12.8 (b). So let a ∈ QuotR′ = QuotR,and assume for a contradiction that a /∈ R′ and a−1 /∈ R′. Then R′[a] and R′[a−1] strictlycontain R′, and thus by maximality of R′ in M we conclude that x ∈ R′[a] and x ∈ R′[a−1]. Sowe can write

x =n

∑i=0

riai =m

∑i=0

sia−i

for some n,m∈N and r0, . . . ,rn,s0, . . . ,sm ∈R′. We can choose n and m minimal, and assumewithout loss of generality that m≤ n. Then

x = s0 +(x− s0)x−1n

∑i=0

riai

= s0 +(x− s0)x−1n−1

∑i=0

riai +m

∑i=1

sia−i x−1 rnan

is a polynomial expression of degree less than n in a with coefficients in R′, in contradictionto the minimality of n. Hence our assumption a /∈ R′ and a−1 /∈ R′ must have been wrong,and we conclude that R′ is a valuation ring. �


Exercise 12.12. Let R be a valuation ring with value group G. Let us call a subset A ⊂ G non-negative if a≥ 0 for all a ∈ G, and saturated if for all a,b ∈ G with a≤ b and a ∈ A we have b ∈ A.Show:

(a) There is a natural inclusion-preserving one-to-one correspondence between ideals of R andnon-negative saturated subsets of G.

(b) For any two ideals I,J ⊂ R we have I ⊂ J or J ⊂ I.

So far we have considered arbitrary valuation rings. However, as most rings occurring in practiceare Noetherian, we now want to specialize to this case. Surprisingly, this seemingly small additionalassumption has far-reaching consequences: it fixes the value group to be Z, restricts the possibilitiesto the 1-dimensional regular local rings of Lemma 12.1, and leads to many other nice properties asthe following two propositions show.

Proposition and Definition 12.13 (Discrete valuation rings). For a valuation ring R with uniquemaximal ideal P (see Lemma 12.6 (d)) the following statements are equivalent:

(a) R is Noetherian, but not a field.

(b) R is a principal ideal domain, but not a field.

(c) The value group of R is Z.

Moreover, in this case the valuation ν(a) ∈ Z of an element a ∈ R\{0} is the unique natural numbern such that (a) = Pn.

If the above equivalent conditions hold, we say that R is a discrete valuation ring (short: DVR).

Proof.

(a)⇒ (b): As R is Noetherian, every ideal I in R can be written as I = (a1, . . . ,ar) for somea1, . . . ,ar by Lemma 7.4 (c). Now if ai has minimal valuation among these elements thena j ∈ (ai) for all j = 1, . . . ,r by Lemma 12.6 (b), and hence I = (ai).

(b)⇒ (c): The maximal ideal P must be non-zero as otherwise R would be a field. So byassumption and Example 2.6 (a) it is of the form P=(t) for a prime element t ∈R. Localizingat (t) again does not change R, and so R = R(t) is the localization of the unique factorizationdomain R (see Example 8.3 (a)) at the ideal (t). By Example 12.4 (c) it is thus a valuationring with value group Z, which shows (c).

Moreover, every non-zero element a ∈ R can be written uniquely as a = ctn for some n ∈ Nand c ∈ R\P. Then c is a unit by Lemma 12.6 (c) and (d), and hence (a) = (tn) = Pn.Example 12.4 (c) shows that ν(a) = n in this case, proving the additional statement of theproposition.

(c)⇒ (a): It is clear that R cannot be a field, since otherwise its valuation group would be(QuotR)∗/R∗ = {1} by Construction 12.7.

Now let IER be a non-zero ideal. As the value group of R is Z, there is a non-zero elementa ∈ I with minimal valuation ν(a). But then I = (a), since for every element b ∈ I we haveν(b)≥ ν(a), and thus b ∈ (a) by Lemma 12.6 (b). �

Proposition 12.14 (Equivalent conditions for discrete valuation rings). For a Noetherian local ringR the following statements are equivalent:

(a) R is a discrete valuation ring.

(b) R is a principal ideal domain, but not a field.

(c) R is a 1-dimensional unique factorization domain.

(d) R is a 1-dimensional normal domain.

(e) R is a 1-dimensional regular ring.


Proof.

(a)⇒ (b) follows from Proposition 12.13.

(b)⇒ (c) holds by Example 8.3 (a) and Example 11.3 (c).

(c)⇒ (d) is Example 9.10.

(d)⇒ (e): As R is a local 1-dimensional domain, its maximal ideal P and the zero ideal are theonly prime ideals in R. So if we choose a ∈ P\{0} then

√(a) = P by Lemma 2.21, and

thus Pn ⊂ (a) for some n ∈ N by Exercise 7.22 (b). Let us choose n minimal. Note thatcertainly n > 0 since otherwise a would have to be a unit. So Pn−1 6⊂ (a), and we can pickb ∈ Pn−1\(a).We set t = a

b ∈ QuotR and claim that t ∈ R and P = (t). To see this, note first that

1t

P =ba

P⊂ 1a

Pn ⊂ 1a(a)⊂ R,

so 1t P is an ideal in R. If it was contained in the maximal ideal P, the R-module homomor-

phism P→P, x 7→ 1t x would satisfy a monic polynomial with coefficients in R by Proposition

3.25. Thus 1t ∈ QuotR would be integral over R. As R is assumed to be normal, this would

imply 1t ∈ R and hence b = 1

t a ∈ (a), in contradiction to our choice of b. So the ideal 1t P is

not contained in the unique maximal ideal P, which means that 1t P = R. In other words we

have P = (t), so P is principal and hence R is a regular 1-dimensional ring by Lemma 11.36(a).

(e)⇒ (a): By Lemma 12.1 we know that the maximal ideal of R is generated by one element t,and that every a ∈ R\{0} can be written as a unit times tn for some n ∈ N. Consequently,every element a ∈ QuotR\{0} can be written as a unit times tn for n ∈ Z. But then a ∈ R(if n ≥ 0) or a−1 ∈ R (if n ≤ 0), and thus R is a valuation ring. It is discrete by Proposition12.13 as it is Noetherian by assumption, and not a field since dimR = 1. �

Remark 12.15.

(a) By Proposition 12.14, the 1-dimensional regular local rings considered in Lemma 12.1 areexactly the discrete valuation rings. Moreover, the additional statement in Proposition 12.13shows that the “valuation” constructed in Lemma 12.1 is in fact the valuation in the sense ofDefinition 12.3 and Proposition 12.8.

(b) In this course we have met two conditions on rings that correspond geometrically to somesort of “non-singularity” statement: normality in Example 9.11, and regularity in Example11.37. Proposition 12.14 states that these two conditions agree in the 1-dimensional case.One can show however that this is no longer true in higher dimensions: regular local ringsare always normal, but the converse is in general false.

Example 12.16.

(a) As in Remark 12.2, every ring of local functions at a smooth point on a curve is a 1-dimensional regular local ring, and thus by Proposition 12.14 a discrete valuation ring. ByRemark 12.15 (a) its valuation can be interpreted as orders of vanishing as in Remark 12.2.

(b) The power series ring L[[t]] over a field L is a discrete valuation ring by Proposition 12.13,since it is a valuation ring with valuation group Z by Example 12.4 (d).

Our results above allow to give a very easy description of the ideals in a discrete valuation ring.

Corollary 12.17 (Ideals in a discrete valuation ring). Let R be a discrete valuation ring with uniquemaximal ideal P. Then the non-zero ideals of R are exactly the powers Pn for n ∈ N. They form astrictly decreasing chain

R = P0 ) P1 ) P2 ) · · · .


Proof. By Proposition 12.13, any non-zero ideal in R is principal and of the form Pn for some n∈N.Moreover, these powers form a strict chain of ideals since Pn = Pn+1 = P ·Pn for some n wouldimply Pn = 0 by Nakayama’s Lemma as in Example 6.16 (a), and thus the contradiction P = 0. �

Finally, the valuation in a discrete valuation ring can also be expressed in terms of the length ofmodules introduced in Definition 3.18.

Corollary 12.18 (Valuation and length). Let R be a discrete valuation ring, and let a∈ R\{0}. Thenν(a) = lR(R/(a)).

Proof. By Proposition 12.13 (b) the maximal ideal P of R can be generated by one element t. More-over, Proposition 12.13 shows that (a) = Pν(a) = (tν(a)). Now

0 = (tν(a))/(a) ( (tν(a)−1)/(a) ( · · · ( (t0)/(a) = R/(a) (∗)is a chain of submodules of R/(a) with successive quotients (t i−1)/(t i) for i = 1, . . . ,ν(a) by Propo-sition 3.10 (b). But

R/(t)→ (t i−1)/(t i), c 7→ ct i−1

is an isomorphism for all i, and thus (∗) is a composition series for R/(a) as an R-module. Hencewe conclude that lR(R/(a)) = ν(a). �

13. Dedekind Domains 117

13. Dedekind Domains

In the last chapter we have mainly studied 1-dimensional regular local rings, i. e. geometrically thelocal properties of smooth points on curves. We now want to patch these local results together toobtain global statements about 1-dimensional rings (resp. curves) that are “locally regular”. Thecorresponding notion is that of a Dedekind domain.

Definition 13.1 (Dedekind domains). An integral domain R is called Dedekind domain if it isNoetherian of dimension 1, and for all maximal ideals PER the localization RP is a regular localring.

Remark 13.2 (Equivalent conditions for Dedekind domains). As a Dedekind domain R is an integraldomain of dimension 1, its prime ideals are exactly the zero ideal and all maximal ideals. So everylocalization RP for a maximal ideal P is a 1-dimensional local ring. As these localizations are alsoNoetherian by Exercise 7.23, we can replace the requirement in Definition 13.1 that the local ringsRP are regular by any of the equivalent conditions in Proposition 12.14. For example, a Dedekinddomain is the same as a 1-dimensional Noetherian domain such that all localizations at maximalideals are discrete valuation rings.

This works particularly well for the normality condition as this is a local property and can thus betransferred to the whole ring:

Lemma 13.3. A 1-dimensional Noetherian domain is a Dedekind domain if and only if it is normal.

Proof. By Remark 13.2 and Proposition 12.14, a 1-dimensional Noetherian domain R is a Dedekinddomain if and only if all localizations RP at a maximal ideal P are normal. But by Exercise 9.13 (c)this is equivalent to R being normal. �

25

Example 13.4.(a) By Lemma 13.3, any principal ideal domain which is not a field is a Dedekind domain: it is

1-dimensional by Example 11.3 (c), clearly Noetherian, and normal by Example 9.10 sinceit is a unique factorization domain by Example 8.3 (a). For better visualization, the followingdiagram shows the implications between various properties of rings for the case of integraldomains that are not fields. Rings that are always 1-dimensional and / or local are markedas such. It is true that every regular local ring is a unique factorization domain, but we havenot proven this here since this requires more advanced methods — we have only shown inProposition 11.40 that any regular local ring is an integral domain.

=⇒

=⇒

=⇒

13.4 (a)

DVR

PID

regular

=⇒

=⇒12.14 (e)

12.14 (b)

local

local

dim = 1

dim = 1Dedekind

UFD

normal domain=⇒

=⇒

=⇒

8.3 (a)

9.10

dim = 1

13.3

def.

(b) Let X be an irreducible curve over an algebraically closed field. Assume that X is smooth,i. e. that all points of X are smooth in the sense of Example 11.37 and Definition 11.38. Thenthe coordinate ring A(X) is a Dedekind domain: it is an integral domain by Lemma 2.3 (a)since I(X) is a prime ideal by Remark 2.7 (b). It is also 1-dimensional by assumption and


Noetherian by Remark 7.15. Moreover, by Hilbert’s Nullstellensatz as in Remark 10.11 themaximal ideals in A(X) are exactly the ideals of points, and so our smoothness assumptionis the same as saying that all localizations at maximal ideals are regular.

In fact, irreducible smooth curves over algebraically closed fields are the main geometricexamples for Dedekind domains. However, there is also a large class of examples in numbertheory, which explains why the concept of a Dedekind domain is equally important in num-ber theory and geometry: it turns out that the ring of integral elements in a number field, i. e.in a finite field extension of Q, is always a Dedekind domain. Let us prove this now.

Proposition 13.5 (Integral elements in number fields). Let Q⊂ K be a finite field extension, and letR be the integral closure of Z in K. Then R is a Dedekind domain.

Proof. As a subring of a field, R is clearly an integral domain. Moreover, by Example 11.3 (c) andLemma 11.8 we have dimR = dimZ= 1. It is also easy to see that R is normal: if a ∈QuotR⊂ K isintegral over R it is also integral over Z by transitivity as in Lemma 9.6 (b), so it is contained in theintegral closure R of Z in K. Hence by Lemma 13.3 it only remains to show that R is Noetherian —which is in fact the hardest part of the proof. We will show this in three steps.

(a) We claim that |R/pR|< ∞ for all prime numbers p ∈ Z.

Note that R/pR is a vector space over Zp = Z/pZ. It suffices to show that dimZp R/pR ≤dimQ K since this dimension is finite by assumption. So let a1, . . . ,an ∈ R/pR be linearlyindependent over Zp. We will show that a1, . . . ,an ∈ K are also independent over Q, so thatn ≤ dimQ K. Otherwise there are λ1, . . . ,λn ∈ Q not all zero with λ1a1 + · · ·+ λnan = 0.After multiplying these coefficients with a common scalar we may assume that all of themare integers, and not all of them are divisible by p. But then λ1 a1 + · · ·+ λn an = 0 isa non-trivial relation in R/pR with coefficients in Zp, in contradiction to a1, . . . ,an beingindependent over Zp.

(b) We will show that |R/mR|< ∞ for all m ∈ Z\{0}.In fact, this follows by induction on the number of prime factors in m: for one prime factorthe statement is just that of (a), and for more prime factors it follows from the exact sequenceof Abelian groups

0−→ R/m1R·m2−→ R/m1m2R−→ R/m2R−→ 0,

since this means that |R/m1m2R|= |R/m1R| · |R/m2R|< ∞.

(c) Now let IER be any non-zero ideal. We claim that m ∈ I for some m ∈ Z\{0}.Otherwise we would have

dimR/I = dimZ/(I∩Z) = dimZ= 1

by Lemma 11.8, since R/I is integral over Z/(I ∩Z) by Lemma 9.7 (a). But dimR has tobe bigger than dimR/I, since a chain of prime ideals in R/I corresponds to a chain of primeideals in R containing I, which can always be extended to a longer chain by the zero idealsince R is an integral domain. Hence dimR > 1, a contradiction.

Putting everything together, we can choose a non-zero m ∈ I ∩Z by (c), so that mRE I. Hence|I/mR| ≤ |R/mR| < ∞ by (b), so I/mR = {a1, . . . ,an} for some a1, . . . ,an ∈ I. But then the idealI = (a1, . . . ,an,m) is finitely generated, and hence R is Noetherian. �

Example 13.6. Consider again the ring R = Z[√

5 i] of Example 8.3 (b). By Example 9.16, it is theintegral closure of Z in Q(

√5 i). Hence Proposition 13.5 shows that R is a Dedekind domain.

We see from this example that a Dedekind domain is in general not a unique factorization domain,as e. g. by Example 8.3 (b) the element 2 is irreducible, but not prime in R, so that it does not havea factorization into prime elements. However, we will prove now that a Dedekind domain alwayshas an analogue of the unique factorization property for ideals, i. e. every non-zero ideal can bewritten uniquely as a product of non-zero prime ideals (which are then also maximal since Dedekind


domains are 1-dimensional). In fact, this is the most important property of Dedekind domains inpractice.

Proposition 13.7 (Prime factorization of ideals in Dedekind domains). Let R be a Dedekind domain.

(a) Let PER be a maximal ideal, and let QER be any ideal. Then

Q is P-primary ⇔ Q = Pk for some k ∈ N>0.

Moreover, the number k is unique in this case.

(b) Any non-zero ideal IER has a “prime factorization”

I = Pk11 · · · · ·P

knn

with k1, . . . ,kn ∈ N>0 and distinct maximal ideals P1, . . . ,PnER. It is unique up to permuta-tion of the factors, and P1, . . . ,Pn are exactly the associated prime ideals of I.

Proof.

(a) The implication “⇐” holds in arbitrary rings by Lemma 8.12 (b), so let us show the oppositedirection “⇒”. Let Q be P-primary, and consider the localization map R→ RP. Then Qe isa non-zero ideal in the localization RP, which is a discrete valuation ring by Remark 13.2.So by Corollary 12.17 we have Qe = (Pe)k for some k, and hence Qe = (Pk)e as extensioncommutes with products by Exercise 1.19 (c). Contracting this equation now gives Q = Pk

by Lemma 8.33, since Q and Pk are both P-primary by Lemma 8.12 (b).

The number k is unique since Pk = Pl for k 6= l would imply (Pe)k = (Pe)l by extension, incontradiction to Corollary 12.17.

(b) As R is Noetherian, the ideal I has a minimal primary decomposition I = Q1 ∩ ·· · ∩Qn byCorollary 8.21. Since I is non-zero, the corresponding associated prime ideals P1, . . . ,Pn ofthese primary ideals are distinct and non-zero, and hence maximal as dimR= 1. In particular,there are no strict inclusions among the ideals P1, . . . ,Pn, and thus all of them are minimalover I. By Proposition 8.34 this means that the ideals Q1, . . . ,Qn in our decomposition areunique.

Now by (a) we have Qi = Pkii for unique ki ∈ N>0 for i = 1, . . . ,n. This gives us a unique

decomposition I = Pk11 ∩ ·· ·∩Pkn

n , and thus also a unique factorization I = Pk11 · · · · ·Pkn

n byExercise 1.8 since the ideals Pk1

1 , . . . ,Pknn are pairwise coprime by Exercise 2.24. �

Example 13.8. Recall from Examples 8.3 (b) and 13.6 that the element 2 in the Dedekind domainR =Z[

√5 i] does not admit a factorization into prime elements. But by Proposition 13.7 (b) the ideal

(2) must have a decomposition as a product of maximal ideals (which cannot all be principal, asotherwise we would have decomposed the number 2 into prime factors). Concretely, we claim thatthis decomposition is

(2) = (2,1+√

5 i)2.

To see this, note first that the ideal (2,1+√

5 i) is maximal by Lemma 2.3 (b) since the quotient

Z[√

5 i]/(2,1+√

5 i) ∼= Z/(2) ∼= Z2

is a field. Moreover, we have (2)⊂ (2,1+√

5 i)2 since

2 = (1+√

5 i)2−22−2√

5 i(1+√

5 i) ∈ (2,1+√

5 i)2,

and (2,1+√

5 i)2 ⊂ (2) as

22 ∈ (2), 2(1+√

5 i) ∈ (2), and (1+√

5 i)2 =−4+2√

5 i ∈ (2).

To understand the geometric meaning of the prime factorization of ideals we need a lemma first.

Lemma 13.9 (Ideals in Dedekind domains). Let R be a Dedekind domain.


(a) For all distinct maximal ideals P1, . . . ,Pn of R and k1, . . . ,kn, l1, . . . , ln ∈ N we have

Pk11 · · · · ·P

knn ⊂ Pl1

1 · · · · ·Plnn ⇔ li ≤ ki for all i = 1, . . . ,n.

(b) For any a ∈ R\{0} we have

(a) = Pν1(a)1 · · · · ·Pνn(a)

n ,

where P1, . . . ,Pn are the associated prime ideals of (a), and νi denotes the valuation of thediscrete valuation ring RPi (restricted to R).

Proof. By Exercise 6.29 (a) ideal containment is a local property, i. e. we can check it on all local-izations RP for maximal ideals P. Moreover, products commute with localization by Exercise 1.19(c), and the localization of Pi at a maximal ideal P 6= Pi is the unit ideal by Example 6.25 (a). Hence:

(a) We have

Pk11 · · · · ·P

knn ⊂ Pl1

1 · · · · ·Plnn ⇔ (Pe

i )ki ⊂ (Pe

i )li in RPi for all i

⇔ li ≤ ki for all i. (Corollary 12.17)

(b) By Proposition 13.7 (b) we know that (a) = Pk11 · · · · · Pkn

n for suitable k1, . . . ,kn ∈ N ifP1, . . . ,Pn are the associated prime ideals of (a). To determine the exponent ki for i= 1, . . . ,n,we localize at RPi to get (a) = (Pe

i )ki in the discrete valuation ring RPi , and use Proposition

12.13 to conclude from this that ki = νi(a). �

Remark 13.10. If a is a non-zero element in a Dedekind domain R and PER a maximal ideal that isnot an associated prime ideal of (a), the same argument as in the proof of Lemma 13.9 (b) shows thatthe valuation of a in the discrete valuation ring RP is 0. Hence the ideals P1, . . . ,Pn in the statementof this lemma are exactly the maximal ideals of R so that the valuation of a in the correspondingdiscrete valuation ring is non-zero.

Remark 13.11 (Geometric interpretation of the prime factorization of ideals). Let X be an irre-ducible smooth curve over an algebraically closed field, so that its coordinate ring R = A(X) is aDedekind domain by Example 13.4 (b). Now let f ∈ R be a non-zero polynomial function on X , andlet a1, . . . ,an ∈ X be the zeroes of f , with corresponding maximal ideals P1, . . . ,PnER. Moreover,for i = 1, . . . ,n let ki be the order of vanishing of f at ai as in Remark 12.2, i. e. the valuation of f inthe ring RPi of local functions on X at ai. Then Lemma 13.9 (b) (together with Remark 13.10) statesthat

( f ) = Pk11 · · · · ·P

knn .

In other words, the prime factorization of a principal ideal ( f ) in the coordinate ring of X encodesthe orders of vanishing of the function f at all points of X . Here is a concrete example of thisconstruction that will also be used later on in Example 13.29.

Example 13.12. Consider the complex plane cubic curve X = V (y2− x(x− 1)(x− λ )) ⊂ A2C for

some λ ∈ C\{0,1}. The picture (a) below shows approximately the real points of X in the caseλ ∈ R>1: the vertical line x = c for c ∈ R intersects X in two real points symmetric with respect tothis axis if 0 < c < 1 or c > λ , in exactly the point (c,0) if c ∈ {0,1,λ}, and in no real point in allother cases.


X

(a) (b)

P2

P1x0 λ1

x = c

lP3

l′

P′1

P′2

It is easy to check as in Example 11.39 (b) that all points of X are smooth, so that the coordinate ringR = C[x,y]/(y2− x(x−1)(x−λ )) of X is a Dedekind domain by Example 13.4 (b).

Now let l ∈ R be a general linear function as in picture (b) above. On the curve X it vanishesat three points (to order 1) since the cubic equation y2 = x(x− 1)(x− λ ) together with a generallinear equation in x and y will have three solutions. If P1, P2, and P3 are the maximal ideals in Rcorresponding to these points, Remark 13.11 shows that (l) = P1 ·P2 ·P3 in R.

Note that for special linear functions it might happen that some of these points coincide, as in the caseof l′ above which vanishes to order 2 at the point P′1. Consequently, in this case we get (l′) = P′1

2 ·P2.

For computational purposes, the unique factorization property for ideals allows us to perform calcu-lations with ideals in Dedekind domains very much in the same way as in principal ideal domains.For example, the following Proposition 13.13 is entirely analogous (both in its statement and in itsproof) to Example 1.4.

Proposition 13.13 (Operations on ideals in Dedekind domains). Let I and J be two non-zero idealsin a Dedekind domain, with prime factorizations

I = Pk11 · · · · ·P

knn and J = Pl1

1 · · · · ·Plnn

as in Proposition 13.7, where P1, . . . ,Pn are distinct maximal ideals and k1, . . . ,kn, l1, . . . , ln ∈ N.Then

I + J = Pm11 · · · · ·P

mnn with mi = min(ki, li),

I∩ J = Pm11 · · · · ·P

mnn with mi = max(ki, li),

I · J = Pm11 · · · · ·P

mnn with mi = ki + li

for i = 1, . . . ,n. In particular, I · J = (I + J) · (I∩ J).

Proof. By Proposition 13.7 (b) we can write all three ideals as Pm11 · · · · ·Pmn

n for suitable m1, . . . ,mn(if we possibly enlarge the set of maximal ideals occurring in the factorizations). So it only remainsto determine the numbers m1, . . . ,mn for the three cases.

The ideal I + J is the smallest ideal containing both I and J. By Lemma 13.9 (a) this means thatmi is the biggest number less than or equal to both ki and li, i. e. min(ki, li). Analogously, theintersection I ∩ J is the biggest ideal contained in both I and J, so in this case mi is the smallestnumber greater than or equal to both ki and li, i. e. max(ki, li). The exponents mi = ki + li for theproduct are obvious. �

As a Dedekind domain is in general not a unique factorization domain, it clearly follows from Exam-ple 8.3 (a) that it is usually not a principal ideal domain either. However, a surprising result followingfrom the computational rules in Proposition 13.13 is that every ideal in a Dedekind domain can begenerated by two elements. In fact, there is an even stronger statement:

Proposition 13.14. Let R be a Dedekind domain, and let a be a non-zero element in an ideal IER.Then there is an element b ∈ R such that I = (a,b).

In particular, every ideal in R can be generated by two elements.


Proof. By assumption (a)⊂ I are non-zero ideals, so we know by Proposition 13.7 (b) and Lemma13.9 (a) that

(a) = Pk11 · · · · ·P

knn and I = Pl1

1 · · · · ·Plnn

for suitable distinct maximal ideals P1, . . . ,PnER and natural numbers li ≤ ki for all i = 1, . . . ,n. Bythe uniqueness part of Proposition 13.7 (b) we can pick elements

bi ∈ Pl1+11 · · · · ·Pli

i · · · · ·Pln+1n

∖Pl1+1

1 · · · · ·Pli+1i · · · · ·Pln+1

n ⊂ I

for all i. Then bi ∈ Pl j+1j for all j 6= i, but bi /∈ Pli+1

i , since otherwise by Proposition 13.13

bi ∈ Pli+1i ∩

(Pl1+1

1 · · · · ·Plii · · · · ·P

ln+1n

)= Pl1+1

1 · · · · ·Pln+1n

in contradiction to our choice of bi. Hence

b := b1 + · · ·+bn /∈ Pli+1i

for all i, but certainly b ∈ I. We now claim that I = (a,b). To see this, note first that by Proposition13.13 the prime factorization of (a,b) = (a)+(b) can contain at most the maximal ideals occurringin (a), so we can write (a,b) = Pm1

1 · · · · ·Pmnn for suitable m1, . . . ,mn. But by Lemma 13.9 (a) we

see that:

• li ≤ mi for all i since (a,b)⊂ I;

• mi ≤ li for all i since b /∈ Pli+1i , and hence (a,b) 6⊂ Pli+1

i .

Therefore we get mi = li for all i, which means that I = (a,b). �26

We have now studied prime factorizations of ideals in Dedekind domains in some detail. However,recall that the underlying valuations on the local rings are defined originally not only on these dis-crete valuation rings, but also on their quotient field. Geometrically, this means that we can equallywell consider orders of rational functions, i. e. quotients of polynomials, at a smooth point of acurve. These orders can then be positive (if the function has a zero), negative (if it has a pole), orzero (if the function has a non-zero value at the given point). Let us now transfer this extensionto the quotient field to the global case of a Dedekind domain R. Instead of ideals we then have toconsider corresponding structures (i. e. R-submodules) that do not lie in R itself, but in its quotientfield QuotR.

Definition 13.15 (Fractional ideals). Let R be an integral domain with quotient field K = QuotR.

(a) A fractional ideal of R is an R-submodule I of K such that aI ⊂ R for some a ∈ R\{0}.(b) For a1, . . . ,an ∈ K we set as expected

(a1, . . . ,an) := Ra1 + · · ·+Ran

and call this the fractional ideal generated by a1, . . . ,an (note that this is in fact a fractionalideal since we can take for a in (a) the product of the denominators of a1, . . . ,an).

Example 13.16.(a) A subset I of an integral domain R is a fractional ideal of R if and only if it is an ideal in R

(the condition in Definition 13.15 (a) that aI ⊂ R for some a is vacuous in this case since wecan always take a = 1).

(b)( 1

2

)= 1

2 Z ⊂ Q is a fractional ideal of Z. In contrast, the localization Z(2) ⊂ Q of Example6.5 (d) is not a fractional ideal of Z: it is a Z-submodule of Q, but there is no non-zerointeger a such that aZ(2) ⊂ Z.

Remark 13.17. Let R be an integral domain with quotient field K.

(a) Let I be a fractional ideal of R. The condition aI ⊂ R of Definition 13.15 (a) ensures thatI is finitely generated if R is Noetherian: as aI is an R-submodule in R it is actually anideal in R, and hence of the form aI = (a1, . . . ,an) for some a1, . . . ,an ∈ R. But then alsoI =

( a1a , . . . ,

ana

)is finitely generated.


(b) The standard operations on ideals of Construction 1.1 can easily be extended to fractionalideals, or more generally to R-submodules of K. In the following, we will mainly needproducts and quotients: for two R-submodules I and J of K we set

IJ :={ n

∑i=1

ai bi : n ∈ N, a1, . . . ,an ∈ I, b1, . . . ,bn ∈ J},

I : K J := {a ∈ K : aJ ⊂ I}.

Note that IJ is just the smallest R-submodule of K containing all products ab for a ∈ Iand b ∈ J, as expected. The index K in the notation of the quotient I : K J distinguishesthis construction from the ordinary ideal quotient I : J = {a ∈ R : aJ ⊂ I}— note that bothquotients are defined but different in general if both I and J are ordinary ideals in R.

Exercise 13.18. Let K be the quotient field of a Noetherian integral domain R. Prove that for anytwo fractional ideals I and J of R and any multiplicatively closed subset S⊂ R we have:

(a) S−1(IJ) = S−1I ·S−1J,

(b) S−1(I : K J) = S−1I : K S−1J.

Our goal in the following will be to check whether the multiplication as in Remark 13.17 (b) definesa group structure on the set of all non-zero fractional ideals of an integral domain R. As associativityand the existence of the neutral element R are obvious, the only remaining question is the existenceof inverse elements, i. e. whether for a given non-zero fractional ideal I there is always anotherfractional ideal J with IJ = R. We will see now that this is indeed the case for Dedekind domains,but not in general integral domains.

Definition 13.19 (Invertible and principal ideals). Let R be an integral domain, and let I be anR-submodule of K = QuotR.

(a) I is called an invertible ideal or (Cartier) divisor if there is an R-submodule of J of K suchthat IJ = R.

(b) I is called principal if I = (a) for some a ∈ K.

Lemma 13.20. As above, let K be the quotient field of an integral domain R, and let I be an R-submodule of K.

(a) If I is an invertible ideal with IJ = R, then J = R : K I.

(b) We have the implications

I non-zero principal ⇒ I invertible ⇒ I fractional.

Proof.

(a) By definition of the quotient, IJ = R implies R = IJ ⊂ I (R : K I) ⊂ R, so we have equalityIJ = I (R : K I). Multiplication by J now gives the desired result J = R : K I.

(b) If I = (a) is principal with a ∈ K∗ then (a) ·( 1

a

)= R, hence I is invertible.

Now let I be an invertible ideal, i. e. I (R : K I) = R by (a). This means thatn

∑i=1

ai bi = 1

for some n ∈ N and a1, . . . ,an ∈ I and b1, . . . ,bn ∈ R : K I. Then we have for all b ∈ I

b =n

∑i=1

ai bi b︸︷︷︸∈R

.

So if we let a ∈ R be the product of the denominators of a1, . . . ,an, we get ab ∈ R. ThereforeaI ⊂ R, i. e. I is fractional. �


Example 13.21.(a) Let R be a principal ideal domain. Then every non-zero fractional ideal I of R is principal:

we have aI ⊂ R for some a ∈ QuotR\{0}. This is an ideal in R, so of the form (b) for someb ∈ R\{0}. It follows that I =

( ba

), i. e. I is principal.

In particular, Lemma 13.20 (b) implies that the notions of principal, invertible, and fractionalideals all agree for non-zero ideals in a principal ideal domain.

(b) The ideal I = (x,y) in the ring R = R[x,y] is not invertible: setting K = QuotR = R(x,y) wehave

R : K I = { f ∈ R(x,y) : x f ∈ R[x,y] and y f ∈ R[x,y]}= R[x,y].

But I (R : K I) = (x,y)R[x,y] 6= R, and hence I is not invertible by Lemma 13.20 (a).

Proposition 13.22 (Invertible = fractional ideals in Dedekind domains). In a Dedekind domain,every non-zero fractional ideal is invertible.

Proof. Let I be a non-zero fractional ideal of a Dedekind domain R. Assume that I is not invertible,which means by Lemma 13.20 (a) that I (R : K I) 6= R. As the inclusion I (R : K I)⊂ R is obvious, thismeans that I (R : K I) is a proper ideal of R. It must therefore be contained in a maximal ideal P byCorollary 2.17.

Extending this inclusion by the localization map R→ RP then gives Ie (RP : K Ie) ⊂ Pe by Exercise13.18. This means by Lemma 13.20 (a) that Ie is not invertible in RP. But RP is a discrete valuationring by Remark 13.2, hence a principal ideal domain by Proposition 12.14, and so Ie cannot be afractional ideal either by Example 13.21 (a). This is clearly a contradiction, since I is assumed to befractional. �

Remark 13.23. By construction, the invertible ideals of an integral domain R form an Abeliangroup under multiplication, with neutral element R. As expected, we will write the inverse R : K I ofan invertible ideal I as in Lemma 13.20 (a) also as I−1. Proposition 13.22 tells us that for Dedekinddomains this group of invertible ideals can also be thought of as the group of non-zero fractionalideals.

Moreover, it is obvious that the non-zero principal fractional ideals form a subgroup:

• every non-zero principal fractional ideal is invertible by Lemma 13.20 (b);

• the neutral element R = (1) is principal;

• for two non-zero principal fractional ideals (a) and (b) their product (ab) is principal;

• for any non-zero principal fractional ideal (a) its inverse (a−1) is also principal.

So we can define the following groups that are naturally attached to any integral domain.

Definition 13.24 (Ideal class groups). Let R be an integral domain.

(a) The group of all invertible ideals of R (under multiplication) is called the ideal group orgroup of (Cartier) divisors of R. We denote it by DivR.

(b) We denote by PrinR≤ DivR the subgroup of (non-zero) principal ideals.

(c) The quotient PicR := DivR/PrinR of all invertible ideals modulo principal ideals is calledthe ideal class group, or group of (Cartier) divisor classes, or Picard group of R.

Let us restrict the study of these groups to Dedekind domains. In this case, the structure of the idealgroup is easy to understand with the following proposition.

Proposition 13.25 (Prime factorization for invertible ideals). Let I be an invertible ideal in aDedekind domain R. Then I = Pk1

1 · · · · · Pknn for suitable distinct maximal ideals P1, . . . ,Pn and

k1, . . . ,kn ∈ Z, and this representation is unique up to permutation of the factors.


Proof. By Lemma 13.20 (b) we know that aI is an ideal in R for a suitable a ∈ R\{0}. Now byProposition 13.7 (b) we have aI = Pr1

1 · · · · ·Prnn and (a) = Ps1

1 · · · · ·Psnn for suitable distinct maximal

ideals P1, . . . ,Pn and r1, . . . ,rn,s1, . . . ,sn ∈ N, and so we get a factorization

I = (a)−1 ·aI = Pr1−s11 · · · · ·Prn−sn

n

as desired. Moreover, if we have two such factorizations Pk11 · · · · · Pkn

n = Pl11 · · · · · Pln

n withk1, . . . ,kn, l1, . . . , ln ∈ Z, we can multiply this equation with suitable powers of P1, . . . ,Pn so thatthe exponents become non-negative. The uniqueness statement then follows from the correspondingone in Proposition 13.7 (b). �

Remark 13.26. Let R be a Dedekind domain. Proposition 13.25 states that the ideal group DivR isin fact easy to describe: we have an isomorphism

DivR→{ϕ : mSpecR→ Z : ϕ is non-zero only on finitely many maximal ideals}

sending any invertible ideal Pk11 · · · · ·Pkn

n to the map ϕ : mSpecR→ Z with only non-zero valuesϕ(Pi) = ki for i = 1, . . . ,n. This is usually called the free Abelian group generated by mSpecR(since the maximal ideals generate this group, and there are no non-trivial relations among thesegenerators).

The group DivR is therefore very “big”, and also at the same time not very interesting since itsstructure is so simple. In contrast, the ideal class group PicR=DivR/PrinR is usually much smaller,and contains a lot of information on R. It is of great importance both in geometry and number theory.It is out of the scope of this course to study it in detail, but we will at least give one interestingexample in each of these areas. But first let us note that the ideal class group can be thought of asmeasuring “how far away R is from being a principal ideal domain”:

Proposition 13.27. For a Dedekind domain R the following statements are equivalent:

(a) R is a principal ideal domain.

(b) R is a unique factorization domain.

(c) PicR is the trivial group, i. e. |PicR|= 1.

Proof.

(a)⇒ (b) is Example 8.3 (a).

(b)⇒ (c): By Exercise 8.32 (b) every maximal ideal PER (which is also a minimal non-zeroprime ideal as dimR = 1) is principal. But these maximal ideals generate the group DivR byProposition 13.25, and so we have PrinR = DivR, i. e. |PicR|= 1.

(c)⇒ (a): By Definition 13.24 the assumption |PicR| = 1 means that every invertible ideal isprincipal. But every non-zero ideal of R is invertible by Proposition 13.22, so the resultfollows. �

Example 13.28 (A non-trivial Picard group in number theory). Let R = Z[√

5 i] be the integralclosure of Z in K = Q(

√5 i) as in Examples 8.3 (b) and 13.6. We have seen there already that R is

a Dedekind domain but not a principal ideal domain: the ideal I1 := (2,1+√

5 i) is not principal. Inparticular, the class of I1 in the Picard group PicR is non-trivial. We will now show that this is theonly non-trivial element in PicR, i. e. that |PicR|= 2 and thus necessarily PicR∼= Z/2Z as a group,with the two elements given by the classes of the ideals I0 := (1) and I1. Unwinding Definition13.24, we therefore claim that every invertible ideal of R is of the form aI0 or aI1 for some a ∈ K∗.

So let I be an invertible ideal of R. Then I is also a non-zero fractional ideal by Lemma 13.20 (b),and thus there is a number b ∈ K∗ with bI ⊂ R. But note that

R = Z[√

5 i] = {m+n√

5 i : m,n ∈ Z}

is just a rectangular lattice in the complex plane, and so there is an element c ∈ bI ⊂ K of minimalnon-zero absolute value. Replacing I by b

c I (which is an equivalent element in the Picard group) we


can therefore assume that I is an invertible ideal with 1 ∈ I, hence R ⊂ I, and that 1 is a non-zeroelement of minimal absolute value in I.Let us find out whether I can contain more elements except the ones of R. To do this, it suffices toconsider points in the rectangle with corners 0, 1,

√5 i, and 1+

√5 i shown in the pictures below:

I is an additive subgroup of C containing R, and so the complete set of points in I will just be anR-periodic repeated copy of the points in this rectangle.To figure out if I contains more points in this rectangle, we proceed in three steps illustrated below.

(a) By construction, I contains no points of absolute value less than 1 except the origin, i. e. nopoints in the open disc U1(0) with radius 1 and center point 0. Likewise, because of theR-periodicity of I, the ideal also does not contain any points in the open unit discs aroundthe other corners of the rectangle, except these corner points themselves. In other words, theshaded area in picture (a) below cannot contain any points of I except the corner points.

√5 i

1

Re

Im √5 i

1

Re

Im √5 i

1

Re

Im

12+

12√

5 i

(a) (b) (c)

(b) Now consider the open disc U 12( 1

2

√5 i), whose intersection with our rectangle is the left dark

half-circle in picture (b) above. Again, it cannot contain any points of I except its center: asI is an R-module, any non-center point a ∈ I in this disc would lead to a non-center point2a ∈ I in the disc U1(

√5 i), which we excluded already in (a). But in fact the center point

12

√5 i cannot lie in I either, since then we would have

√5 i · 1

2

√5 i+3 ·1 =

12∈ I

as well, in contradiction to (a). In the same way we see that the open disc U 12(1+ 1

2

√5 i)

does not contain any points of I either. Hence the complete shaded area in picture (b) isexcluded now for points of I.

(c) Finally, consider the open disc U 12( 1

2 +12

√5 i), shown in picture (c) above in dark color. For

the same reason as in (b), no point in this disc except the center can lie in I. As our discsnow cover the complete rectangle, this means that the only point in our rectangle except thecorners that can be in I is 1

2 +12

√5 i. This leads to exactly two possibilities for I:

either I = R = I0 or I =(

1,12+

12

√5 i)=

12

I1.

In fact, we know already that this last case 12 I1 is an invertible ideal of R, so that this time

(in contrast to (b) above) it does not lead to a contradiction if the center point of the disc liesin I.

Altogether, we thus conclude that |PicR| = 2, i. e. PicR ∼= Z/2Z, with the class of I0 being neutraland I1 being the unique other element. We have indeed also checked already that I1 is its own inversein PicR, since by Example 13.8

I1 · I1 = (2)


is principal, and hence the neutral element in PicR.

Example 13.29 (A non-trivial Picard group in geometry). Consider again the complex plane cubiccurve X = V (y2− x(x− 1)(x− λ )) ⊂ A2

C for some λ ∈ C\{0,1} as in Example 13.12. We havealready seen that its coordinate ring R = A(X) is a Dedekind domain. Let us now study its idealclass group PicR.

Note that there is an obvious map

ϕ : X → PicR, a 7→ I(a)

that assigns to each point of X the class of its maximal ideal in PicR = DivR/PrinR. The surprisingfact is that this map ϕ is injective with its image equal to PicR\{(1)}, i. e. to PicR without its neutralelement. We can therefore make it bijective by adding a “point at infinity” to X that is mapped tothe missing point (1) of PicR. This is particularly interesting as we then have a bijection betweentwo completely different algebraic structures: X is a variety (but a priori not a group) and PicR is agroup (but a priori not a variety). So we can use the bijection ϕ to make the cubic curve X ∪{∞}into a group, and the group PicR into a variety.

We cannot prove this statement or study its consequences here sincethis would require methods that we have not covered in this course— this is usually done in the “Algebraic Geometry” class. However,we can use our definition of the Picard group and the map ϕ aboveto describe the group structure on the curve X ∪{∞} explicitly. Todo this, consider two points on the curve with maximal ideals P1and P2, as shown in the picture on the right. Draw the line throughthese two points; it will intersect X in one more point Q since thecubic equation y2 = x(x−1)(x−λ ) together with a linear equationin x and y will have three solutions. By Example 13.12, this meansalgebraically that there is a linear polynomial l ∈ R (whose zerolocus is this line) such that (l) = P1 ·P2 ·Q.

X

P2

P1

Q

P

ll′

Next, draw the vertical line through Q. By the symmetry of X , it will intersect X in one more pointP. Similarly to the above, it follows that there is a linear polynomial l′ such that (l′) = Q ·P. But inthe Picard group PicR this means that

P1 ·P2 ·Q = (l) = (l′) = Q ·P(note that (l) and (l′) both define the neutral element in PicR), and thus that P1 ·P2 = P. Hencethe above geometric construction of P from P1 and P2 describes the group structure on X ∪ {∞}mentioned above.

Note that it is quite obvious without much theory behind it that this geometric two-line constructioncan be used to associate to any two points on X (corresponding to P1 and P2 above) a third point onX (corresponding to P). The surprising statement here (which is very hard to prove without usingPicard groups) is that this gives rise to a group structure on X ∪{∞}; in particular that this operationis associative. The neutral element is the additional point ∞ (corresponding to the class of principalideals in PicR), and the inverse of a point in X is the other intersection point of the vertical linethrough this point with X (so that e. g. in the above picture we have P−1 = Q).

27

128 References

References

[AM] M. Atiyah, I. MacDonald, Introduction to Commutative Algebra, Addison Wesley (2004)[E] D. Eisenbud, Commutative Algebra with a View towards Algebraic Geometry, Springer (2004)[G1] A. Gathmann, Algebraische Strukturen, class notes TU Kaiserslautern (2019/20),

https://www.mathematik.uni-kl.de/~gathmann/ags[G2] A. Gathmann, Grundlagen der Mathematik, class notes TU Kaiserslautern (2018/19),

https://www.mathematik.uni-kl.de/~gathmann/gdm[G3] A. Gathmann, Einführung in die Algebra, class notes TU Kaiserslautern (2010),

https://www.mathematik.uni-kl.de/~gathmann/algebra[HM] S. Hampe, T. Markwig, Commutative Algebra, class notes TU Kaiserslautern (2012),

www.math.uni-tuebingen.de/~keilen/download/Lehre/MGSS09/CommutativeAlg.pdf[K] T. Keilen, Algebraic Structures, class notes TU Kaiserslautern (2009),

www.math.uni-tuebingen.de/~keilen/download/LectureNotes/algebraicstructures.pdf

[M] T. Markwig, Elementare Zahlentheorie, class notes TU Kaiserslautern (2010),www.math.uni-tuebingen.de/user/keilen/download/LectureNotes/zahlentheorie.pdf

[S] W. Decker, G.-M. Greuel, G. Pfister, H. Schönemann, SINGULAR — a computer algebra system for polynomialcomputations,https://www.singular.uni-kl.de

https://www.mathematik.uni-kl.de/~gathmann/ags

https://www.mathematik.uni-kl.de/~gathmann/gdm

https://www.mathematik.uni-kl.de/~gathmann/algebra

www.math.uni-tuebingen.de/~keilen/download/Lehre/MGSS09/CommutativeAlg.pdf



www.math.uni-tuebingen.de/user/keilen/download/LectureNotes/zahlentheorie.pdf

https://www.singular.uni-kl.de

Index 129

Index

5-Lemma 41

AnK 4

adjoint matrix 31affine space 4affine variety 4algebra 15

finitely generated 16tensor product 49

algebra homomorphism 15algebraic element 80algebraically closed field 11algebraically dependent 103algebraically independent 103annihilator 30annM 30antisymmetric relation 21Artinian module 62Artinian ring 62

Structure Theorem 69ascending chain condition 62Ass(I) 77associated prime ideal 77Axiom of Choice 23

basisof a finitely generated module 31transcendence 103

Basis Theorem 66bilinear map 44bound 21

Cartier divisor 123class group 124group 124

Cayley-Hamilton 34chain condition

ascending 62descending 62

Chinese Remainder Theorem 12class group

divisor 124ideal 124

codimP 96codimension

of a prime ideal 96of an irreducible subvariety 96

commutative diagram 15complexification 48composition series 32connecting homomorphism 38contraction

of an ideal 13coordinate ring 5coprime ideals 10curve 96

Dedekind domain 117

degreeof a polynomial 8of transcendence 104

dependentalgebraically 103

descending chain condition 62determinant 31diagram

commutative 15dimR 96dimension

of a ring 96of a variety 96

direct sumof submodules 28

discrete valuation ring 114DivR 124divisor

Cartier 123class group 124group 124

domain 8Dedekind 117Euclidean 10factorial 70integral 8normal 83principal ideal 10unique factorization 70

dual vector space 47DVR 114

elementalgebraic 80integral 80irreducible 19maximal 21prime 19

embedded prime ideal 77Euclidean domain 10exact sequence 36

gluing 37localization 58short 36split 42splitting 37

extensionof an ideal 13of scalars 48

extension ring 80finite 80integral 80

factorial ring 70fiber

of a morphism 81field 3

130 Index

algebraically closed 11field extension

finite 80finite field extension 80finite ring extension 80finitely generated Abelian group

Structure Theorem 65finitely generated algebra 16finitely generated module 28First Uniqueness Theorem

for primary decompositions 78fractional ideal 122

invertible 123principal 123

free module 31function

local 52polynomial 5

Gauß 71generated subalgebra 16generated submodule 28gluing exact sequences 37Going Down 89Going Up 88group

ideal 124ideal class 124of divisor classes 124of divisors 124ordered 110Picard 124

heightof a prime ideal 96

Hilbert’s Basis Theorem 66Hilbert’s Nullstellensatz 93–95Hom(M,N) 28

left exact 40homomorphism

connecting 38localization 57of algebras 15of modules 28

Hopkins 68

I(X) 5ideal

class group 124contraction 13coprime 10extension 13fractional 122group 124image 13inverse image 13invertible 123maximal 18of a variety 5P-primary 74primary 73prime 18principal 10, 123product 9

quotient 9radical 9symbolic power 101

imageof a module homomorphism 29

image ideal 13Incomparability 87independent

algebraically 103inert prime ideal 103integral closure 83integral domain 8

factorial 70normal 83

integral element 80integral ring extension 80integrally closed 83intersection

of ideals 9inverse image ideal 13invertible ideal 123irreducible element 19irreducible variety 19isolated prime ideal 77isomorphic modules 28isomorphism

of modules 28isomorphism theorems 29

K[x1, . . . ,xn] 3K(x1, . . . ,xn) 103kernel

of a module homomorphism 29Krull

dimension 96Principal Ideal Theorem 101

Laurent series 111left exact 40lemma

5- 41of Gauß 71of Nakayama 34, 56of Zorn 21, 22Snake 38Splitting 41

lengthof a module 32

linear map 28local function 52local ring 56

regular 107localization

at a multiplicatively closed set 53at a prime ideal 54at an element 54exactness 58of a homomorphism 57of a module 57of a ring 53universal property 56

Lying Over 86

matrix

Index 131

adjoint 31maximal element 21maximal ideal 18maximal spectrum 18minimal polynomial 84minimal primary decomposition 76minimal prime ideal 25module 27

Artinian 62direct sum 28finitely generated 28free 31homomorphism 28isomorphic 28isomorphism 28Noetherian 62quotient 28, 30sum 28tensor product 44

monic polynomial 34, 80monomial tensor 45morphism

of algebras 15of modules 28of varieties 6

mSpecR 18multiplicatively closed set 53

saturation 56

Nakayama’s Lemma 34, 56nilpotent 9nilradical 9Noether Normalization 92Noetherian module 62Noetherian ring 21, 62normal domain 83normalization

Noether 92Nullstellensatz 93–95

orderpartial 21total 21

ordered group 110

partial order 21PicR 124Picard group 124PID 10polynomial 3

minimal 84monic 34, 80

polynomial function 5polynomial ring 3P-primary ideal 74primary decomposition 74

First Uniqueness Theorem 78minimal 76Second Uniqueness Theorem 79

primary ideal 73prime element 19prime ideal 18

associated 77embedded 77

inert 103isolated 77minimal 25ramified 103split 103

PrinR 124principal fractional ideal 123principal ideal 10, 123principal ideal domain 10Principal Ideal Theorem 101product

of ideals 9tensor 44

Puiseux series 111pure tensor 45

quotientof ideals 9of modules 28, 30

quotient field 54QuotR 54

Ra 54RP 54R[x1, . . . ,xn] 3radical

of an ideal 9radical ideal 9ramified prime ideal 103rank

of a free module 31reduced ring 9reducible variety 19reflexive 21regular local ring 107regular point 107relation

antisymmetric 21reflexive 21transitive 21

right exact 50ring 3

Artinian 62discrete valuation 114factorial 70integrally closed 83local 56Noetherian 21, 62of functions 5of local functions 54of polynomial functions 5of polynomials 3reduced 9regular local 107valuation 110

ring extension 80finite 80integral 80

S−1R 53saturation

of a multiplicatively closed set 56Second Uniqueness Theorem

for primary decompositions 79

132 Index

sequenceexact 36split exact 42

seriesLaurent 111Puiseux 111

setpartially ordered 21totally ordered 21well-ordered 23

short exact sequence 36singular point 107smooth point 107Snake Lemma 38space

affine 4SpecR 18spectrum 18

maximal 18split exact sequence 42split prime ideal 103splitting exact sequence 37Splitting Lemma 41Structure Theorem

for Artinian rings 69for finitely generated Abelian groups 65

subalgebra 15generated 16

submodule 28generated 28

subvariety 5sum

of ideals 9of submodules 28

symbolic power 101

tangent space 105tensor 44

monomial 45pure 45

tensor productof algebras 49of homomorphisms 47of modules 44right exact 50universal property 44

total order 21transcendence basis 103transcendence degree 104transitive 21trivial valuation 110

UFD 70unique factorization domain 70Uniqueness Theorem for primary decomposition

first 78second 79

unit 19universal property

of localization 56of tensor products 44

upper bound 21

V (S) 4

valuation 109, 110trivial 110value group 110

valuation ring 110discrete 114

valueof a polynomial 4

value group 110variety

affine 4dimension 96irreducible 19reducible 19

vector spacedual 47

well-order 23

zero locus 4zero-divisor 8Zorn’s Lemma 21, 22